gr-gifted-amateur6-2-cdfb94c6-a223-4235-b28e-3935c3dbbd28

Embedding

When a blind beetle crawls over the surface of a curved branch, it doesn't notice that the track it has covered is indeed curved. I was lucky enough to notice what the beetle didn't notice.
Albert Einstein
In everyday life, if an object shows some degree of curvature (like the surface of a bowl, for example) we know about the curvature because the object is embedded in the three-dimensional Euclidean space of our everyday experience. 1 1 ^(1){ }^{1}1 Mathematically, we can describe the geometry of the two-dimensional surface of the bowl using the three coordinates of the Euclidean space. It might therefore seem that a natural way to describe curved spacetime would be to embed it in a higher dimensional Euclidean space. However, there is absolutely no evidence that this is how Nature works and so, to avoid introducing unnecessary structure to the geometry of spacetime, we seek a way to describe curved space using only the coordinates available to a geometer confined to that space. Our fate, as geometers of a spacetime that we cannot step outside of is, therefore, a little like that of a beetle walking on the curved surface of the bowl. If the beetle is unable to escape the surface, then the coordinates it uses to measure lengths between points on the bowl will be two-dimensional. Despite this, we can use the notion of embedding to (i) work out the metric of a curved space in terms of those coordinates that a trapped geometer would use, and (ii) given a metric in terms of the coordinates within the surface, to visualize the space by embedding it in a Euclidean space. In order to embed a space, you need more than one more dimension in the Euclidean space than are used in the object's metric. 2 2 ^(2){ }^{2}2
In order to tackle embedding, we start by addressing a related problem. Say we already have an object in Euclidean space. We can then work out the metric on its surface in terms of the coordinates confined within that surface. We do this by relating the Euclidean coordinates to the internal, or object coordinates, and the expression for the metric we then obtain, in terms of the object's coordinates, is known as an induced metric.
Recall that the distance between points in Euclidean 3 -space R 3 R 3 R^(3)\mathbb{R}^{3}R3 with coordinates X α = ( X , Y , Z ) X α = ( X , Y , Z ) X^(alpha)=(X,Y,Z)X^{\alpha}=(X, Y, Z)Xα=(X,Y,Z) is given by
(D.1) d s 2 = d X 2 + d Y 2 + d Z 2 (D.1) d s 2 = d X 2 + d Y 2 + d Z 2 {:(D.1)ds^(2)=dX^(2)+dY^(2)+dZ^(2):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} X^{2}+\mathrm{d} Y^{2}+\mathrm{d} Z^{2} \tag{D.1} \end{equation*}(D.1)ds2=dX2+dY2+dZ2
Chapter summary 585
Exercises
1 1 ^(1){ }^{1}1 Our discussion in this Appendix follows Zee, which can be consulted for further details.
2 2 ^(2){ }^{2}2 John Nash (1928-2015) is perhaps most famous for his Nobel Memorial Prize-winning work in Economics. His work in geometry includes the Nash embedding theorems, which demonstrate that any Riemannian manifold can be embedded in a Euclidean space. Silvia Nasar's biography of Nash, A Beautiful Mind, comes highly recommended, but should not be confused with the film of the same name, with which it bears little resemblance.
Fig. D. 1 Circle embedded in R 2 R 2 R^(2)\mathbb{R}^{2}R2.
This expression generalizes to
(D.2) d s 2 = α = 1 N ( d X α ) 2 (D.2) d s 2 = α = 1 N d X α 2 {:(D.2)ds^(2)=sum_(alpha=1)^(N)((d)X^(alpha))^(2):}\begin{equation*} \mathrm{d} s^{2}=\sum_{\alpha=1}^{N}\left(\mathrm{~d} X^{\alpha}\right)^{2} \tag{D.2} \end{equation*}(D.2)ds2=α=1N( dXα)2
for N N NNN-dimensional Euclidean space R N R N R^(N)\mathbb{R}^{N}RN. In what follows, we shall embed objects from D D DDD-dimensional spaces into R N R N R^(N)\mathbb{R}^{N}RN. There are two methods to consider to find the induced metric.
Method I I III is suitable for when we have an equation for each of the Euclidean coordinates X α X α X^(alpha)X^{\alpha}Xα in terms of the coordinates of the D D DDD-dimensional object x μ x μ x^(mu)x^{\mu}xμ. We write the coordinates X α ( x 1 x D ) X α x 1 x D X^(alpha)(x^(1)dotsx^(D))X^{\alpha}\left(x^{1} \ldots x^{D}\right)Xα(x1xD), where α = 1 , , N α = 1 , , N alpha=1,dots,N\alpha=1, \ldots, Nα=1,,N. We start by saying
(D.3) X α + d X α = X α + X α x μ d x μ (D.3) X α + d X α = X α + X α x μ d x μ {:(D.3)X^(alpha)+dX^(alpha)=X^(alpha)+(delX^(alpha))/(delx^(mu))dx^(mu):}\begin{equation*} X^{\alpha}+\mathrm{d} X^{\alpha}=X^{\alpha}+\frac{\partial X^{\alpha}}{\partial x^{\mu}} \mathrm{d} x^{\mu} \tag{D.3} \end{equation*}(D.3)Xα+dXα=Xα+Xαxμdxμ
Then we use eqn D. 1 to say
d s 2 = α ( d X α ) 2 = α X α x μ d x μ X α x ν d x ν (D.4) = g μ ν d x μ d x ν d s 2 = α d X α 2 = α X α x μ d x μ X α x ν d x ν (D.4) = g μ ν d x μ d x ν {:[ds^(2)=sum_(alpha)(dX^(alpha))^(2)],[=sum_(alpha)(delX^(alpha))/(delx^(mu))dx^(mu)(delX^(alpha))/(delx^(nu))dx^(nu)],[(D.4)=g_(mu nu)dx^(mu)dx^(nu)]:}\begin{align*} \mathrm{d} s^{2} & =\sum_{\alpha}\left(\mathrm{d} X^{\alpha}\right)^{2} \\ & =\sum_{\alpha} \frac{\partial X^{\alpha}}{\partial x^{\mu}} \mathrm{d} x^{\mu} \frac{\partial X^{\alpha}}{\partial x^{\nu}} \mathrm{d} x^{\nu} \\ & =g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} \tag{D.4} \end{align*}ds2=α(dXα)2=αXαxμdxμXαxνdxν(D.4)=gμνdxμdxν
where, in the final line, we've noted the general rule that d s 2 = d s 2 = ds^(2)=\mathrm{d} s^{2}=ds2= g μ ν d x μ d x ν g μ ν d x μ d x ν g_(mu nu)dx^(mu)dx^(nu)g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}gμνdxμdxν. We deduce that the relationship between the metric in the object space g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν is related to the embedding coordinates via
(D.5) g μ ν = α X α x μ X α x ν (D.5) g μ ν = α X α x μ X α x ν {:(D.5)g_(mu nu)=sum_(alpha)(delX^(alpha))/(delx^(mu))(delX^(alpha))/(delx^(nu)):}\begin{equation*} g_{\mu \nu}=\sum_{\alpha} \frac{\partial X^{\alpha}}{\partial x^{\mu}} \frac{\partial X^{\alpha}}{\partial x^{\nu}} \tag{D.5} \end{equation*}(D.5)gμν=αXαxμXαxν
Method II is suitable for use when you can express the shape of the object you're trying to represent in Euclidean coordinates. Again we start with the metric line element in Euclidean space d s 2 = d X 2 + d s 2 = d X 2 + ds^(2)=dX^(2)+\mathrm{d} s^{2}=\mathrm{d} X^{2}+ds2=dX2+ d Y 2 + d Z 2 d Y 2 + d Z 2 dY^(2)+dZ^(2)\mathrm{d} Y^{2}+\mathrm{d} Z^{2}dY2+dZ2. We then write an equation describing the object Z = f ( X , Y ) Z = f ( X , Y ) Z=f(X,Y)Z=f(X, Y)Z=f(X,Y) in terms of the embedding coordinates, which allows us to eliminate a coordinate Z Z ZZZ. We then require an educated guess at some good objectspace coordinates x μ x μ x^(mu)x^{\mu}xμ to reduce the metric to something convenient.

Example D. 1

(a) Consider a circle. If we confine ourselves to the surface of the circle it is a onedimensional space, so D = 1 D = 1 D=1D=1D=1. The coordinate of particular points on the circle can be specified by giving an object coordinate θ θ theta\thetaθ.
Start with method I. We can mount a circle in Euclidean ( N = 2 N = 2 N=2N=2N=2 )-space (see Fig. D.1) by writing
(D.6) X = a cos θ and Y = a sin θ (D.6) X = a cos θ  and  Y = a sin θ {:(D.6)X=a cos thetaquad" and "quad Y=a sin theta:}\begin{equation*} X=a \cos \theta \quad \text { and } \quad Y=a \sin \theta \tag{D.6} \end{equation*}(D.6)X=acosθ and Y=asinθ
where a a aaa is the radius of the circle in the Euclidean space. We have therefore written a relationship between the circle's coordinate θ θ theta\thetaθ and the embedding coordinates X α X α X^(alpha)X^{\alpha}Xα. Differentiating, we have
(D.7) X θ = a sin θ and Y θ = a cos θ (D.7) X θ = a sin θ  and  Y θ = a cos θ {:(D.7)(del X)/(del theta)=-a sin thetaquad" and "quad(del Y)/(del theta)=a cos theta:}\begin{equation*} \frac{\partial X}{\partial \theta}=-a \sin \theta \quad \text { and } \quad \frac{\partial Y}{\partial \theta}=a \cos \theta \tag{D.7} \end{equation*}(D.7)Xθ=asinθ and Yθ=acosθ
This enables us to read out the metric component using eqn D. 5 as
(D.8) g θ θ = a 2 sin 2 θ + a 2 cos 2 θ = a 2 (D.8) g θ θ = a 2 sin 2 θ + a 2 cos 2 θ = a 2 {:(D.8)g_(theta theta)=a^(2)sin^(2)theta+a^(2)cos^(2)theta=a^(2):}\begin{equation*} g_{\theta \theta}=a^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta=a^{2} \tag{D.8} \end{equation*}(D.8)gθθ=a2sin2θ+a2cos2θ=a2
and we have the result
(D.9) d s 2 = g θ θ d θ d θ = a 2 d θ 2 (D.9) d s 2 = g θ θ d θ d θ = a 2 d θ 2 {:(D.9)ds^(2)=g_(theta theta)dthetadtheta=a^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=g_{\theta \theta} \mathrm{d} \theta \mathrm{~d} \theta=a^{2} \mathrm{~d} \theta^{2} \tag{D.9} \end{equation*}(D.9)ds2=gθθdθ dθ=a2 dθ2
Now try method II. We write
(D.10) d s 2 = d X 2 + d Y 2 = d X 2 [ 1 + ( Y X ) 2 ] (D.10) d s 2 = d X 2 + d Y 2 = d X 2 1 + Y X 2 {:(D.10)ds^(2)=dX^(2)+dY^(2)=dX^(2)[1+((del Y)/(del X))^(2)]:}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} X^{2}+\mathrm{d} Y^{2}=\mathrm{d} X^{2}\left[1+\left(\frac{\partial Y}{\partial X}\right)^{2}\right] \tag{D.10} \end{equation*}(D.10)ds2=dX2+dY2=dX2[1+(YX)2]
The equation for the circle in embedding coordinates is Y = ( a 2 X 2 ) 1 2 Y = a 2 X 2 1 2 Y=(a^(2)-X^(2))^((1)/(2))Y=\left(a^{2}-X^{2}\right)^{\frac{1}{2}}Y=(a2X2)12, so we have
(D.11) Y X = X ( a 2 X 2 ) 1 2 (D.11) Y X = X a 2 X 2 1 2 {:(D.11)(del Y)/(del X)=(-X)/((a^(2)-X^(2))^((1)/(2))):}\begin{equation*} \frac{\partial Y}{\partial X}=\frac{-X}{\left(a^{2}-X^{2}\right)^{\frac{1}{2}}} \tag{D.11} \end{equation*}(D.11)YX=X(a2X2)12
leading to
(D.12) d s 2 = d X 2 ( 1 + X 2 ( a 2 X 2 ) ) (D.12) d s 2 = d X 2 1 + X 2 a 2 X 2 {:(D.12)ds^(2)=dX^(2)(1+(X^(2))/((a^(2)-X^(2)))):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} X^{2}\left(1+\frac{X^{2}}{\left(a^{2}-X^{2}\right)}\right) \tag{D.12} \end{equation*}(D.12)ds2=dX2(1+X2(a2X2))
Now, if we let X = a cos θ X = a cos θ X=a cos thetaX=a \cos \thetaX=acosθ and d X = a sin θ d θ d X = a sin θ d θ dX=-a sin thetadtheta\mathrm{d} X=-a \sin \theta \mathrm{~d} \thetadX=asinθ dθ, we find (as before)
(D.13) d s 2 = a 2 d θ 2 (D.13) d s 2 = a 2 d θ 2 {:(D.13)ds^(2)=a^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=a^{2} \mathrm{~d} \theta^{2} \tag{D.13} \end{equation*}(D.13)ds2=a2 dθ2
(b) Let's now go up a dimension and embed the unit D = 2 D = 2 D=2D=2D=2 sphere in R 3 R 3 R^(3)\mathbb{R}^{3}R3 (see Fig. D.2). We need two object-space coordinates ( x 1 , x 2 ) = ( θ , ϕ ) x 1 , x 2 = ( θ , ϕ ) (x^(1),x^(2))=(theta,phi)\left(x^{1}, x^{2}\right)=(\theta, \phi)(x1,x2)=(θ,ϕ). We know a parametrization in R 3 R 3 R^(3)\mathbb{R}^{3}R3 using spherical coordinates with r = 1 r = 1 r=1r=1r=1 we write ( X , Y , Z ) = ( X , Y , Z ) = (X,Y,Z)=(X, Y, Z)=(X,Y,Z)= ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) ( sin θ cos ϕ , sin θ sin ϕ , cos θ ) (sin theta cos phi,sin theta sin phi,cos theta)(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)(sinθcosϕ,sinθsinϕ,cosθ). We plug in to find the metric induced by the embedding
g θ θ = ( X θ ) 2 + ( Y θ ) 2 + ( Z θ ) 2 (D.14) = cos 2 θ cos 2 ϕ + cos 2 θ sin 2 ϕ + sin 2 θ = 1 g ϕ ϕ = ( X ϕ ) 2 + ( Y ϕ ) 2 + ( Z ϕ ) 2 = sin 2 θ (D.15) = sin 2 θ sin 2 ϕ + sin 2 θ cos 2 ϕ = sin 2 θ g θ θ = X θ 2 + Y θ 2 + Z θ 2 (D.14) = cos 2 θ cos 2 ϕ + cos 2 θ sin 2 ϕ + sin 2 θ = 1 g ϕ ϕ = X ϕ 2 + Y ϕ 2 + Z ϕ 2 = sin 2 θ (D.15) = sin 2 θ sin 2 ϕ + sin 2 θ cos 2 ϕ = sin 2 θ {:[g_(theta theta)=((del X)/(del theta))^(2)+((del Y)/(del theta))^(2)+((del Z)/(del theta))^(2)],[(D.14)=cos^(2)thetacos^(2)phi+cos^(2)thetasin^(2)phi+sin^(2)theta=1],[g_(phi phi)=((del X)/(del phi))^(2)+((del Y)/(del phi))^(2)+((del Z)/(del phi))^(2)=sin^(2)theta],[(D.15)=sin^(2)thetasin^(2)phi+sin^(2)thetacos^(2)phi=sin^(2)theta]:}\begin{align*} g_{\theta \theta} & =\left(\frac{\partial X}{\partial \theta}\right)^{2}+\left(\frac{\partial Y}{\partial \theta}\right)^{2}+\left(\frac{\partial Z}{\partial \theta}\right)^{2} \\ & =\cos ^{2} \theta \cos ^{2} \phi+\cos ^{2} \theta \sin ^{2} \phi+\sin ^{2} \theta=1 \tag{D.14}\\ g_{\phi \phi} & =\left(\frac{\partial X}{\partial \phi}\right)^{2}+\left(\frac{\partial Y}{\partial \phi}\right)^{2}+\left(\frac{\partial Z}{\partial \phi}\right)^{2}=\sin ^{2} \theta \\ & =\sin ^{2} \theta \sin ^{2} \phi+\sin ^{2} \theta \cos ^{2} \phi=\sin ^{2} \theta \tag{D.15} \end{align*}gθθ=(Xθ)2+(Yθ)2+(Zθ)2(D.14)=cos2θcos2ϕ+cos2θsin2ϕ+sin2θ=1gϕϕ=(Xϕ)2+(Yϕ)2+(Zϕ)2=sin2θ(D.15)=sin2θsin2ϕ+sin2θcos2ϕ=sin2θ
and find g ϕ θ = 0 g ϕ θ = 0 g_(phi theta)=0g_{\phi \theta}=0gϕθ=0. We conclude that the D = 2 D = 2 D=2D=2D=2 sphere is described by the induced line element
(D.16) d s 2 = d θ 2 + sin 2 θ d ϕ 2 . (D.16) d s 2 = d θ 2 + sin 2 θ d ϕ 2 . {:(D.16)ds^(2)=dtheta^(2)+sin^(2)thetadphi^(2).:}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2} . \tag{D.16} \end{equation*}(D.16)ds2=dθ2+sin2θ dϕ2. Let's now try method II. The equation of the spherical surface in R 3 R 3 R^(3)\mathbb{R}^{3}R3 is X 2 + Y 2 + X 2 + Y 2 + X^(2)+Y^(2)+X^{2}+Y^{2}+X2+Y2+ Z 2 = 1 Z 2 = 1 Z^(2)=1Z^{2}=1Z2=1. Eliminating Z Z ZZZ we have
(D.17) d s 2 = d X 2 + d Y 2 + ( X d X + Y d Y ) 2 1 X 2 Y 2 (D.17) d s 2 = d X 2 + d Y 2 + ( X d X + Y d Y ) 2 1 X 2 Y 2 {:(D.17)ds^(2)=dX^(2)+dY^(2)+((X(d)X+Y(d)Y)^(2))/(1-X^(2)-Y^(2)):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} X^{2}+\mathrm{d} Y^{2}+\frac{(X \mathrm{~d} X+Y \mathrm{~d} Y)^{2}}{1-X^{2}-Y^{2}} \tag{D.17} \end{equation*}(D.17)ds2=dX2+dY2+(X dX+Y dY)21X2Y2
Now to check for consistency with some well-chosen coordinates. The metric looks cylindrically symmetric, so we replace ( X , Y ) ( X , Y ) (X,Y)(X, Y)(X,Y) with cylindrical coordinates. Use X = X = X=X=X= r cos θ , Y = r sin θ , X 2 + Y 2 = r 2 r cos θ , Y = r sin θ , X 2 + Y 2 = r 2 r cos theta,Y=r sin theta,X^(2)+Y^(2)=r^(2)r \cos \theta, Y=r \sin \theta, X^{2}+Y^{2}=r^{2}rcosθ,Y=rsinθ,X2+Y2=r2 and X d X + Y d Y = r d r X d X + Y d Y = r d r XdX+YdY=rdrX \mathrm{~d} X+Y \mathrm{~d} Y=r \mathrm{~d} rX dX+Y dY=r dr. We find
(D.18) d s 2 = d r 2 1 r 2 + r 2 d ϕ 2 (D.18) d s 2 = d r 2 1 r 2 + r 2 d ϕ 2 {:(D.18)ds^(2)=(dr^(2))/(1-r^(2))+r^(2)dphi^(2):}\begin{equation*} \mathrm{d} s^{2}=\frac{\mathrm{d} r^{2}}{1-r^{2}}+r^{2} \mathrm{~d} \phi^{2} \tag{D.18} \end{equation*}(D.18)ds2=dr21r2+r2 dϕ2

R 3 R 3 R^(3)\mathbb{R}^{3}R3. D. 2 The 2-sphere, embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
This looks different to the method-I answer, until we realize we can write r = sin θ r = sin θ r=sin thetar=\sin \thetar=sinθ, then d r 2 = cos 2 θ d θ 2 d r 2 = cos 2 θ d θ 2 dr^(2)=cos^(2)thetadtheta^(2)\mathrm{d} r^{2}=\cos ^{2} \theta \mathrm{~d} \theta^{2}dr2=cos2θ dθ2 and so d s 2 = d θ 2 + sin 2 θ d ϕ 2 d s 2 = d θ 2 + sin 2 θ d ϕ 2 ds^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} s^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}ds2=dθ2+sin2θ dϕ2 as we had before.
Let's now turn to the problem of visualizing metrics. Often we are faced with the situation where we are presented with a line element and we want to know what the surface looks like when it's embedded in a higher dimensional Euclidean space. The next example shows an example of how to do this for the Einstein-Rosen bridge.
Fig. D. 3 The function Z ( r ) = Z ( r ) = Z(r)=Z(r)=Z(r)= ± [ 4 r S ( r r S ) ] 1 2 ± 4 r S r r S 1 2 +-[4r_(S)(r-r_(S))]^((1)/(2))\pm\left[4 r_{\mathrm{S}}\left(r-r_{\mathrm{S}}\right)\right]^{\frac{1}{2}}±[4rS(rrS)]12.
Fig. D. 4 The wormhole embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3.
3 3 ^(3){ }^{3}3 These regions are called asymptotically flat: they are flat at large distances away from the throat.
Fig. D. 5 (a) The wormhole embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 showing a possible spacetime topology. (b) A topologically identical spacetime, showing the wormhole linking two distant points in spacetime.
4 4 ^(4){ }^{4}4 Like the Alcubierre warp drive from Chapter 5, the creation of a wormhole requires negative energies.

Example D. 2

The Einstein-Rosen bridge (also known as a wormhole) is described by a D = 2 D = 2 D=2D=2D=2 metric
(D.19) d s 2 = r d r 2 r r S + r 2 d ϕ 2 (D.19) d s 2 = r d r 2 r r S + r 2 d ϕ 2 {:(D.19)ds^(2)=(r(d)r^(2))/(r-r_(S))+r^(2)dphi^(2):}\begin{equation*} \mathrm{d} s^{2}=\frac{r \mathrm{~d} r^{2}}{r-r_{\mathrm{S}}}+r^{2} \mathrm{~d} \phi^{2} \tag{D.19} \end{equation*}(D.19)ds2=r dr2rrS+r2 dϕ2
where r S r S r_(S)r_{\mathrm{S}}rS is a constant length. Let's embed this into R 3 R 3 R^(3)\mathbb{R}^{3}R3 using method I. We require
( X 1 r ) 2 + ( X 2 r ) 2 + ( X 3 r ) 2 = g r r = r r r S (D.20) ( X 1 θ ) 2 + ( X 2 θ ) 2 + ( X 3 θ ) 2 = g ϕ ϕ = r 2 X 1 r 2 + X 2 r 2 + X 3 r 2 = g r r = r r r S (D.20) X 1 θ 2 + X 2 θ 2 + X 3 θ 2 = g ϕ ϕ = r 2 {:[((delX^(1))/(del r))^(2)+((delX^(2))/(del r))^(2)+((delX^(3))/(del r))^(2)=g_(rr)=(r)/(r-r_(S))],[(D.20)((delX^(1))/(del theta))^(2)+((delX^(2))/(del theta))^(2)+((delX^(3))/(del theta))^(2)=g_(phi phi)=r^(2)]:}\begin{align*} & \left(\frac{\partial X^{1}}{\partial r}\right)^{2}+\left(\frac{\partial X^{2}}{\partial r}\right)^{2}+\left(\frac{\partial X^{3}}{\partial r}\right)^{2}=g_{r r}=\frac{r}{r-r_{\mathrm{S}}} \\ & \left(\frac{\partial X^{1}}{\partial \theta}\right)^{2}+\left(\frac{\partial X^{2}}{\partial \theta}\right)^{2}+\left(\frac{\partial X^{3}}{\partial \theta}\right)^{2}=g_{\phi \phi}=r^{2} \tag{D.20} \end{align*}(X1r)2+(X2r)2+(X3r)2=grr=rrrS(D.20)(X1θ)2+(X2θ)2+(X3θ)2=gϕϕ=r2
We try a set of coordinates ( X 1 , X 2 , X 3 ) = ( r cos θ , r sin θ , Z ( r ) ) X 1 , X 2 , X 3 = ( r cos θ , r sin θ , Z ( r ) ) (X^(1),X^(2),X^(3))=(r cos theta,r sin theta,Z(r))\left(X^{1}, X^{2}, X^{3}\right)=(r \cos \theta, r \sin \theta, Z(r))(X1,X2,X3)=(rcosθ,rsinθ,Z(r)), which solve the second equation, as can be easily seen. The first equation then becomes
(D.21) 1 + ( Z ( r ) r ) 2 = r r r S (D.21) 1 + Z ( r ) r 2 = r r r S {:(D.21)1+((del Z(r))/(del r))^(2)=(r)/(r-r_(S)):}\begin{equation*} 1+\left(\frac{\partial Z(r)}{\partial r}\right)^{2}=\frac{r}{r-r_{\mathrm{S}}} \tag{D.21} \end{equation*}(D.21)1+(Z(r)r)2=rrrS
which is solved if we set Z ( r ) 2 = 4 r S ( r r S ) Z ( r ) 2 = 4 r S r r S Z(r)^(2)=4r_(S)(r-r_(S))Z(r)^{2}=4 r_{\mathrm{S}}\left(r-r_{\mathrm{S}}\right)Z(r)2=4rS(rrS). The function Z ( r ) Z ( r ) Z(r)Z(r)Z(r) is graphed in Fig. D.3. If it is rotated about the z z zzz-axis it generates the spacetime shown in Fig. D.4. The wormhole metric, embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 appears to show two regions of flat 3 3 ^(3){ }^{3}3 spacetime joined by a throat. (The narrow hole down the middle explains the name wormhole, of course.) The metric tells us about the detailed geometry of spacetime, but tells us nothing about the topology of the spacetime, that is, the large scale shape and us nothing about the topology of the spacetime, that is, the large scale shape and
connectivity of the space. As a result, the wormhole might link two completely connectivity of the space. As a result, the wormhole might link two completely
different parts of a spacetime. One possibility is shown in Fig. D.5(a), where the wormhole links two distant points in a spacetime. A topologically identical diagram is shown in Fig. D.5(b), revealing that two distant patches of flat spacetime are linked by the wormhole. If the distance between the two regions via the flat space is greater than the distance through the throat, then we potentially have a shortcut between very distant points in spacetime. It has also been suggested that the wormhole might joint two disconnected parts of spacetime, allowing an explorer to visit a different Universe entirely. 4 4 ^(4){ }^{4}4
There are some surfaces that cannot be embedded in Euclidean space. The most important example for us is hyperbolic space.

Example D. 3

Consider a space described by a metric
(D.22) d s 2 = d χ 2 + sinh 2 χ d ϕ 2 . (D.22) d s 2 = d χ 2 + sinh 2 χ d ϕ 2 . {:(D.22)ds^(2)=dchi^(2)+sinh^(2)chidphi^(2).:}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} \chi^{2}+\sinh ^{2} \chi \mathrm{~d} \phi^{2} . \tag{D.22} \end{equation*}(D.22)ds2=dχ2+sinh2χ dϕ2.
Using the method-I rule, we need
1 = ( X χ ) 2 + ( Y χ ) 2 + ( Z χ ) 2 (D.23) sinh 2 χ = ( X ϕ ) 2 + ( Y ϕ ) 2 + ( Z ϕ ) 2 1 = X χ 2 + Y χ 2 + Z χ 2 (D.23) sinh 2 χ = X ϕ 2 + Y ϕ 2 + Z ϕ 2 {:[1=((del X)/(del chi))^(2)+((del Y)/(del chi))^(2)+((del Z)/(del chi))^(2)],[(D.23)sinh^(2)chi=((del X)/(del phi))^(2)+((del Y)/(del phi))^(2)+((del Z)/(del phi))^(2)]:}\begin{align*} 1 & =\left(\frac{\partial X}{\partial \chi}\right)^{2}+\left(\frac{\partial Y}{\partial \chi}\right)^{2}+\left(\frac{\partial Z}{\partial \chi}\right)^{2} \\ \sinh ^{2} \chi & =\left(\frac{\partial X}{\partial \phi}\right)^{2}+\left(\frac{\partial Y}{\partial \phi}\right)^{2}+\left(\frac{\partial Z}{\partial \phi}\right)^{2} \tag{D.23} \end{align*}1=(Xχ)2+(Yχ)2+(Zχ)2(D.23)sinh2χ=(Xϕ)2+(Yϕ)2+(Zϕ)2
We can try some coordinates for X X XXX and Y Y YYY that retain the cylindrical symmetry of this space, and determine the Z Z ZZZ coordinate. We try
(D.24) X = sinh χ cos ϕ , Y = sinh χ sin ϕ (D.24) X = sinh χ cos ϕ , Y = sinh χ sin ϕ {:(D.24)X=sinh chi cos phi","quad Y=sinh chi sin phi:}\begin{equation*} X=\sinh \chi \cos \phi, \quad Y=\sinh \chi \sin \phi \tag{D.24} \end{equation*}(D.24)X=sinhχcosϕ,Y=sinhχsinϕ
Plugging in to the second equation, we see that Z / ϕ = 0 Z / ϕ = 0 del Z//del phi=0\partial Z / \partial \phi=0Z/ϕ=0, so Z Z ZZZ is indeed cylindrically symmetric. The first equation tells us that
(D.25) Z χ = ( 1 cosh 2 χ ) 1 2 = ( sinh 2 χ ) 1 2 (D.25) Z χ = 1 cosh 2 χ 1 2 = sinh 2 χ 1 2 {:(D.25)(del Z)/(del chi)=(1-cosh^(2)chi)^((1)/(2))=(-sinh^(2)chi)^((1)/(2)):}\begin{equation*} \frac{\partial Z}{\partial \chi}=\left(1-\cosh ^{2} \chi\right)^{\frac{1}{2}}=\left(-\sinh ^{2} \chi\right)^{\frac{1}{2}} \tag{D.25} \end{equation*}(D.25)Zχ=(1cosh2χ)12=(sinh2χ)12
This equation has no real solutions. The embedding has failed.
This embedding is indeed impossible. There are two options, the first is simply to embed part of the surface. 5 5 ^(5){ }^{5}5 Perhaps a more satisfactory solution is not to mount the space in Euclidean space, but rather in some other space. For hyperbolic spaces, the natural higher dimensional space is known as pseudo-Euclidean space which has a metric
(D.26) d s 2 = d X 2 + d Y 2 d W 2 (D.26) d s 2 = d X 2 + d Y 2 d W 2 {:(D.26)ds^(2)=dX^(2)+dY^(2)-dW^(2):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} X^{2}+\mathrm{d} Y^{2}-\mathrm{d} W^{2} \tag{D.26} \end{equation*}(D.26)ds2=dX2+dY2dW2

Example D. 4

(a) Consider a surface X 2 + Y 2 W 2 = 1 X 2 + Y 2 W 2 = 1 X^(2)+Y^(2)-W^(2)=-1X^{2}+Y^{2}-W^{2}=-1X2+Y2W2=1, known as a 2 -sheet hyperboloid. This can be embedded in pseudo-Euclidean space. 6 6 ^(6){ }^{6}6 Use the coordinates
(D.27) X = sinh χ cos θ , Y = sinh χ sin θ , W = cosh χ (D.27) X = sinh χ cos θ , Y = sinh χ sin θ , W = cosh χ {:(D.27)X=sinh chi cos theta","quad Y=sinh chi sin theta","quad W=cosh chi:}\begin{equation*} X=\sinh \chi \cos \theta, \quad Y=\sinh \chi \sin \theta, \quad W=\cosh \chi \tag{D.27} \end{equation*}(D.27)X=sinhχcosθ,Y=sinhχsinθ,W=coshχ
which are consistent with the equation for the surface. Differentiating, we find that
d s 2 = d X 2 + d Y 2 d W 2 (D.28) = d χ 2 + sinh 2 χ d θ 2 . d s 2 = d X 2 + d Y 2 d W 2 (D.28) = d χ 2 + sinh 2 χ d θ 2 . {:[ds^(2)=dX^(2)+dY^(2)-dW^(2)],[(D.28)=dchi^(2)+sinh^(2)chidtheta^(2).]:}\begin{align*} \mathrm{d} s^{2} & =\mathrm{d} X^{2}+\mathrm{d} Y^{2}-\mathrm{d} W^{2} \\ & =\mathrm{d} \chi^{2}+\sinh ^{2} \chi \mathrm{~d} \theta^{2} . \tag{D.28} \end{align*}ds2=dX2+dY2dW2(D.28)=dχ2+sinh2χ dθ2.
This is the metric we considered before. We conclude that the embedding is possible in the pseudo-Euclidean metric. The surface defined by these coordinates is shown in the Fig. D. 6.
(b) Now try the surface X 2 + Y 2 W 2 = 1 X 2 + Y 2 W 2 = 1 X^(2)+Y^(2)-W^(2)=1X^{2}+Y^{2}-W^{2}=1X2+Y2W2=1, known as the one-sheet hyperboloid. This one is solved for the coordinate choice
(D.29) X = cosh χ cos θ , Y = cosh χ sin θ , W = sinh χ (D.29) X = cosh χ cos θ , Y = cosh χ sin θ , W = sinh χ {:(D.29)X=cosh chi cos theta","quad Y=cosh chi sin theta","quad W=sinh chi:}\begin{equation*} X=\cosh \chi \cos \theta, \quad Y=\cosh \chi \sin \theta, \quad W=\sinh \chi \tag{D.29} \end{equation*}(D.29)X=coshχcosθ,Y=coshχsinθ,W=sinhχ
resulting in the metric
(D.30) d s 2 = d χ 2 + cosh 2 χ d θ 2 (D.30) d s 2 = d χ 2 + cosh 2 χ d θ 2 {:(D.30)ds^(2)=-dchi^(2)+cosh^(2)chidtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} \chi^{2}+\cosh ^{2} \chi \mathrm{~d} \theta^{2} \tag{D.30} \end{equation*}(D.30)ds2=dχ2+cosh2χ dθ2
Interestingly, we can turn this into something more familiar by setting cosh χ = r cosh χ = r cosh chi=r\cosh \chi=rcoshχ=r, from which we find
(D.31) d s 2 = d r 2 1 r 2 + r 2 d θ 2 (D.31) d s 2 = d r 2 1 r 2 + r 2 d θ 2 {:(D.31)ds^(2)=(dr^(2))/(1-r^(2))+r^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=\frac{\mathrm{d} r^{2}}{1-r^{2}}+r^{2} \mathrm{~d} \theta^{2} \tag{D.31} \end{equation*}(D.31)ds2=dr21r2+r2 dθ2
which is simply the metric for the 2 -sphere we had before. In fact, we could have generated this metric directly with the choice of coordinates
(D.32) X = r cos θ , Y = r sin θ , W = ( r 2 1 ) 1 2 (D.32) X = r cos θ , Y = r sin θ , W = r 2 1 1 2 {:(D.32)X=r cos theta","quad Y=r sin theta","quad W=(r^(2)-1)^((1)/(2)):}\begin{equation*} X=r \cos \theta, \quad Y=r \sin \theta, \quad W=\left(r^{2}-1\right)^{\frac{1}{2}} \tag{D.32} \end{equation*}(D.32)X=rcosθ,Y=rsinθ,W=(r21)12
The spherical line element that represents the one-sheet hyperboloid should not be misinterpreted to mean that the shape of the space is a sphere. We have eliminated W W WWW leaving a set of concentric circles behind, with distances encoded in the difference between the circles. (Put another way, we have sliced the spacetime leaving circular cross sections.) If we set r = sin θ r = sin θ r=sin thetar=\sin \thetar=sinθ as we did for the 2 -sphere then we would not be able to solve the equation that defines the surface: r 2 W 2 = 1 r 2 W 2 = 1 r^(2)-W^(2)=1r^{2}-W^{2}=1r2W2=1. Setting r = cosh χ r = cosh χ r=cosh chir=\cosh \chir=coshχ does not cause this problem. See Chapter 18 for more details on this space which is very useful in cosmology. method I to this case.
Fig. D. 6 The 2-sheet hyperboloid embedded in pseudo-Euclidean space.

Chapter summary

  • Embedding allows us to visualize a metric space by mounting it in a higher dimensional Euclidean space.

Exercises

(D.1) A parabolic 2-surface in Euclidean 3-space (see (D.3) The D = 2 D = 2 D=2D=2D=2 torus may be embedded in R 3 R 3 R^(3)\mathbb{R}^{3}R3 (Fig. D.8) Fig. D.7) is described by
(D.33) Z = a 2 ( X 2 + Y 2 ) (D.33) Z = a 2 X 2 + Y 2 {:(D.33)Z=(a)/(2)(X^(2)+Y^(2)):}\begin{equation*} Z=\frac{a}{2}\left(X^{2}+Y^{2}\right) \tag{D.33} \end{equation*}(D.33)Z=a2(X2+Y2)
Using method II, show that this leads to the metric
(D.34) d s 2 = ( 1 + a 2 r 2 ) d r 2 + r 2 d θ 2 (D.34) d s 2 = 1 + a 2 r 2 d r 2 + r 2 d θ 2 {:(D.34)ds^(2)=(1+a^(2)r^(2))dr^(2)+r^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=\left(1+a^{2} r^{2}\right) \mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2} \tag{D.34} \end{equation*}(D.34)ds2=(1+a2r2)dr2+r2 dθ2
We saw in Example D.3 how embedding fails for a hyperbolic surface. However, we can embed part of this surface. Consider an alternative metric for a hyperbolic surface
(D.36) d s 2 = d χ 2 + cosh 2 χ d ϕ 2 (D.36) d s 2 = d χ 2 + cosh 2 χ d ϕ 2 {:(D.36)ds^(2)=dchi^(2)+cosh^(2)chidphi^(2):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} \chi^{2}+\cosh ^{2} \chi \mathrm{~d} \phi^{2} \tag{D.36} \end{equation*}(D.36)ds2=dχ2+cosh2χ dϕ2
Using method I show that we have
(D.37) Z χ = ( 1 sinh 2 χ ) 1 2 (D.37) Z χ = 1 sinh 2 χ 1 2 {:(D.37)(del Z)/(del chi)=(1-sinh^(2)chi)^((1)/(2)):}\begin{equation*} \frac{\partial Z}{\partial \chi}=\left(1-\sinh ^{2} \chi\right)^{\frac{1}{2}} \tag{D.37} \end{equation*}(D.37)Zχ=(1sinh2χ)12
This equation does enjoy real solutions for sinh χ < sinh χ < sinh chi <\sinh \chi<sinhχ< 1 , corresponding to the range 0 χ sinh 1 1 0 χ sinh 1 1 0 <= chi <= sinh^(-1)10 \leq \chi \leq \sinh ^{-1} 10χsinh11. So a partial embedding of this surface is possible. This wasn't the hyperbolic line element we started with, but like that line element, it does have a constant negative curvature. Minding's theorem [named after Ferdinand Minding (1806-1885)J says that all constant-curvature surfaces have the same local geometry, and so the geometry of this line element is representative of the hyperbolic spacetime.
(D.5) Consider the Rindler coordinate system ( x 1 , x 2 ) = x 1 , x 2 = (x^(1),x^(2))=\left(x^{1}, x^{2}\right)=(x1,x2)= ( t , x ) ( t , x ) (t,x)(t, x)(t,x), embedded in a two-dimensional Minkowski space ( T , X ) ( T , X ) (T,X)(T, X)(T,X) via
(D.38) T = x sinh t , X = x cosh t (D.38) T = x sinh t , X = x cosh t {:(D.38)T=x sinh t","quad X=x cosh t:}\begin{equation*} T=x \sinh t, \quad X=x \cosh t \tag{D.38} \end{equation*}(D.38)T=xsinht,X=xcosht
Compute the components of the induced metric. Hint: Since we want to work in Minkowski space, rather than Euclidean space, we must use
(D.39) g μ ν = η α β X α x μ X β x ν (D.39) g μ ν = η α β X α x μ X β x ν {:(D.39)g_(mu nu)=eta_(alpha beta)(delX^(alpha))/(delx^(mu))*(delX^(beta))/(delx^(nu)):}\begin{equation*} g_{\mu \nu}=\eta_{\alpha \beta} \frac{\partial X^{\alpha}}{\partial x^{\mu}} \cdot \frac{\partial X^{\beta}}{\partial x^{\nu}} \tag{D.39} \end{equation*}(D.39)gμν=ηαβXαxμXβxν

Answers to selected problems

(0.1) Equating kinetic energy 1 2 m v esc 2 1 2 m v esc  2 (1)/(2)mv_("esc ")^(2)\frac{1}{2} m v_{\text {esc }}^{2}12mvesc 2 to the gravitational potential energy G M m / r G M m / r GMm//rG M m / rGMm/r yields v esc = 2 G M / r v esc  = 2 G M / r v_("esc ")=sqrt(2GM//r)v_{\text {esc }}=\sqrt{2 G M / r}vesc =2GM/r and since Φ = G M m / r Φ = G M m / r Phi=-GMm//r\Phi=-G M m / rΦ=GMm/r then we also have that v esc = 2 | Φ | v esc  = 2 | Φ | v_("esc ")=sqrt(2|Phi|)v_{\text {esc }}=\sqrt{2|\Phi|}vesc =2|Φ|. Rearranging for v esc = c v esc  = c v_("esc ")=cv_{\text {esc }}=cvesc =c gives the Schwarzschild radius r = 2 G M / c 2 r = 2 G M / c 2 r=2GM//c^(2)r=2 G M / c^{2}r=2GM/c2.
(0.2) (i) Earth: 9.8 m s 2 ; 11.2 km s 1 ( 0.00004 c ) 9.8 m s 2 ; 11.2 km s 1 ( 0.00004 c ) 9.8ms^(-2);11.2kms^(-1)(~~0.00004 c)9.8 \mathrm{~m} \mathrm{~s}^{-2} ; 11.2 \mathrm{~km} \mathrm{~s}^{-1}(\approx 0.00004 c)9.8 m s2;11.2 km s1(0.00004c); (ii) Sun: 274 m s 2 ; 6.2 × 10 5 m s 1 ( 0.002 c ) 274 m s 2 ; 6.2 × 10 5 m s 1 ( 0.002 c ) 274ms^(-2);6.2 xx10^(5)ms^(-1)(~~0.002 c)274 \mathrm{~m} \mathrm{~s}^{-2} ; 6.2 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}(\approx 0.002 c)274 m s2;6.2×105 m s1(0.002c); (iii) neutron star: 1.9 × 10 12 m s 2 ; 1.9 × 10 8 m s 1 ( 0.64 c ) 1.9 × 10 12 m s 2 ; 1.9 × 10 8 m s 1 ( 0.64 c ) 1.9 xx10^(12)ms^(-2);1.9 xx10^(8)ms^(-1)(~~0.64 c)1.9 \times 10^{12} \mathrm{~m} \mathrm{~s}^{-2} ; 1.9 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}(\approx 0.64 c)1.9×1012 m s2;1.9×108 m s1(0.64c).
(0.3) The inverse square law of force in Newtonian gravity implies that | Δ F | / | F | = 2 | Δ r | / | r | | Δ F | / | F | = 2 | Δ r | / | r | |Delta F|//|F|=2|Delta r|//|r||\Delta F| /|F|=2|\Delta r| /|r||ΔF|/|F|=2|Δr|/|r| and so with Δ r = 1.8 m Δ r = 1.8 m Delta r=1.8m\Delta r=1.8 \mathrm{~m}Δr=1.8 m we have that for the Earth the effect of the tidal force is | Δ F | / | F | = 2 × 1.8 m / R | Δ F | / | F | = 2 × 1.8 m / R |Delta F|//|F|=2xx1.8m//R_(o+)|\Delta F| /|F|=2 \times 1.8 \mathrm{~m} / R_{\oplus}|ΔF|/|F|=2×1.8 m/R which is around 0.6 ppm , i.e. the tidal force is less than a millionth of gravitational force, which is why you don't notice it. The Schwarzschild radius for a 3 M 3 M 3M_(o.)3 M_{\odot}3M black hole is about 9 km , so | Δ F | / | F | 0.0004 | Δ F | / | F | 0.0004 |Delta F|//|F|~~0.0004|\Delta F| /|F| \approx 0.0004|ΔF|/|F|0.0004 (though the tidal forces are enormous since the gravitational field is more than eleven orders of magnitude bigger than it is on Earth). The Schwarzschild radius for a than it is on Earth). The Schwarzschild radius for a
10 6 M 10 6 M 10^(6)M_(o.)10^{6} M_{\odot}106M black hole is about three million km , and hence 10 6 M 10 6 M 10^(6)M_(o.)10^{6} M_{\odot}106M black hole is about three million km, and hence
| Δ F | / | F | 4 × 10 10 | Δ F | / | F | 4 × 10 10 |Delta F|//|F|~~4xx10^(-10)|\Delta F| /|F| \approx 4 \times 10^{-10}|ΔF|/|F|4×1010. The tidal forces are smaller in | Δ F | / | F | 4 × 10 10 | Δ F | / | F | 4 × 10 10 |Delta F|//|F|~~4xx10^(-10)|\Delta F| /|F| \approx 4 \times 10^{-10}|ΔF|/|F|4×1010. The tidal forces are smaller in hole is only about six orders of magnitude greater than on Earth.
(At the Schwarzschild radius, since r M r M r prop Mr \propto MrM then | Δ F | | Δ F | |Delta F|prop|\Delta F| \propto|ΔF| F / r F / r F//rF / rF/r and since F M / r 2 F M / r 2 F prop M//r^(2)F \propto M / r^{2}FM/r2 then | Δ F | M 2 | Δ F | M 2 |Delta F|propM^(-2)|\Delta F| \propto M^{-2}|ΔF|M2, so that tidal forces at the Schwarzschild radius are larger for small black holes than for supermassive black holes.)
(1.2) γ = ( 1 0.995 2 ) 1 / 2 10 γ = 1 0.995 2 1 / 2 10 gamma=(1-0.995^(2))^(-1//2)~~10\gamma=\left(1-0.995^{2}\right)^{-1 / 2} \approx 10γ=(10.9952)1/210. (a) 0.995 c × 2.2 μ s = 0.66 km 0.995 c × 2.2 μ s = 0.66 km 0.995 c xx2.2 mus=0.66km0.995 c \times 2.2 \mu \mathrm{~s}=0.66 \mathrm{~km}0.995c×2.2μ s=0.66 km. (b) Multiply 0.66 km by γ : 6.6 km γ : 6.6 km gamma:6.6km\gamma: 6.6 \mathrm{~km}γ:6.6 km.
(1.3) The interval is given by Δ s 2 = c 2 Δ t 2 + Δ x 2 Δ s 2 = c 2 Δ t 2 + Δ x 2 Deltas^(2)=-c^(2)Deltat^(2)+Deltax^(2)\Delta s^{2}=-c^{2} \Delta t^{2}+\Delta x^{2}Δs2=c2Δt2+Δx2. The time interval Δ t Δ t Delta t\Delta tΔt is given by the difference between the time of events q q qqq and p p ppp as measured by us using coordinates ( t , x ) ( t , x ) (t,x)(t, x)(t,x). Take the time coordinate of event r r rrr to be the origin. This means the event p p ppp takes place at a time t p = τ 2 t p = τ 2 t_(p)=tau_(2)t_{p}=\tau_{2}tp=τ2. Since the light takes a time τ 2 + τ 1 τ 2 + τ 1 tau_(2)+tau_(1)\tau_{2}+\tau_{1}τ2+τ1 to reach q q qqq and return, we reason that the event q q qqq takes place at a time t q = ( τ 2 + τ 1 ) / 2 t q = τ 2 + τ 1 / 2 t_(q)=(tau_(2)+tau_(1))//2t_{q}=\left(\tau_{2}+\tau_{1}\right) / 2tq=(τ2+τ1)/2. As a result we have that
Δ t = t q t p = ( τ 2 + τ 1 ) 2 τ 2 (E.1) = ( τ 1 τ 2 ) 2 Δ t = t q t p = τ 2 + τ 1 2 τ 2 (E.1) = τ 1 τ 2 2 {:[Delta t=t_(q)-t_(p)=((tau_(2)+tau_(1)))/(2)-tau_(2)],[(E.1)=((tau_(1)-tau_(2)))/(2)]:}\begin{align*} \Delta t & =t_{q}-t_{p}=\frac{\left(\tau_{2}+\tau_{1}\right)}{2}-\tau_{2} \\ & =\frac{\left(\tau_{1}-\tau_{2}\right)}{2} \tag{E.1} \end{align*}Δt=tqtp=(τ2+τ1)2τ2(E.1)=(τ1τ2)2
In order to work out the interval in position Δ x Δ x Delta x\Delta xΔx, we note that the distance travelled from our world line to
the event q q qqq is measured by the light pulse as Δ x = Δ x = Delta x=\Delta x=Δx= c ( τ 1 + τ 2 ) / 2 c τ 1 + τ 2 / 2 c(tau_(1)+tau_(2))//2c\left(\tau_{1}+\tau_{2}\right) / 2c(τ1+τ2)/2. Plugging into the expression for the interval Δ s 2 Δ s 2 Deltas^(2)\Delta s^{2}Δs2, we obtain
Δ s 2 = c 2 Δ t 2 + Δ x 2 = c 2 ( τ 1 τ 2 ) 2 4 + c 2 ( τ 1 + τ 2 ) 2 4 (E.2) = c 2 τ 1 τ 2 Δ s 2 = c 2 Δ t 2 + Δ x 2 = c 2 τ 1 τ 2 2 4 + c 2 τ 1 + τ 2 2 4 (E.2) = c 2 τ 1 τ 2 {:[Deltas^(2)=-c^(2)Deltat^(2)+Deltax^(2)],[=-(c^(2)(tau_(1)-tau_(2))^(2))/(4)+(c^(2)(tau_(1)+tau_(2))^(2))/(4)],[(E.2)=c^(2)tau_(1)tau_(2)]:}\begin{align*} \Delta s^{2} & =-c^{2} \Delta t^{2}+\Delta x^{2} \\ & =-\frac{c^{2}\left(\tau_{1}-\tau_{2}\right)^{2}}{4}+\frac{c^{2}\left(\tau_{1}+\tau_{2}\right)^{2}}{4} \\ & =c^{2} \tau_{1} \tau_{2} \tag{E.2} \end{align*}Δs2=c2Δt2+Δx2=c2(τ1τ2)24+c2(τ1+τ2)24(E.2)=c2τ1τ2
(1.4) (a) 0.140 , (b) 0.417 , (c) 0.866 , (d) 0.996 , (e) 0.99995 .
(2.1) This is a straightforward application of eqn 2.38 .
(2.2) Write u = γ ( u ) ( 1 , u ) u = γ ( u ) ( 1 , u ) u=gamma( vec(u))(1, vec(u))\boldsymbol{u}=\gamma(\vec{u})(1, \vec{u})u=γ(u)(1,u) and v = γ ( v ) ( 1 , v ) v = γ ( v ) ( 1 , v ) v=gamma( vec(v))(1, vec(v))\boldsymbol{v}=\gamma(\vec{v})(1, \vec{v})v=γ(v)(1,v), so that u v = γ ( u ) γ ( v ) ( 1 + u v ) u v = γ ( u ) γ ( v ) ( 1 + u v ) u*v=gamma( vec(u))gamma( vec(v))(-1+ vec(u)* vec(v))\boldsymbol{u} \cdot \boldsymbol{v}=\gamma(\vec{u}) \gamma(\vec{v})(-1+\vec{u} \cdot \vec{v})uv=γ(u)γ(v)(1+uv) and this is equal to γ γ -gamma-\gammaγ, yielding the result. In the non-relativistic limit, γ = ( 1 v rel 2 ) 1 / 2 = 1 + v rel 2 / 2 + γ = 1 v rel 2 1 / 2 = 1 + v rel 2 / 2 + gamma=(1-v_(rel)^(2))^(-1//2)=1+v_(rel)^(2)//2+cdots\gamma=\left(1-v_{\mathrm{rel}}^{2}\right)^{-1 / 2}=1+v_{\mathrm{rel}}^{2} / 2+\cdotsγ=(1vrel2)1/2=1+vrel2/2+ and hence
(E.3) 1 + v rel 2 2 + = ( 1 + u 2 2 + ) ( 1 + v 2 2 + ) ( 1 u v ) (E.3) 1 + v rel  2 2 + = 1 + u 2 2 + 1 + v 2 2 + ( 1 u v ) {:(E.3)1+(v_("rel ")^(2))/(2)+cdots=(1+(u^(2))/(2)+cdots)(1+(v^(2))/(2)+cdots)(1- vec(u)* vec(v)):}\begin{equation*} 1+\frac{v_{\text {rel }}^{2}}{2}+\cdots=\left(1+\frac{u^{2}}{2}+\cdots\right)\left(1+\frac{v^{2}}{2}+\cdots\right)(1-\vec{u} \cdot \vec{v}) \tag{E.3} \end{equation*}(E.3)1+vrel 22+=(1+u22+)(1+v22+)(1uv)
(E.4) = 1 + ( u v ) 2 2 + (E.4) = 1 + ( u v ) 2 2 + {:(E.4)=1+((( vec(u))-( vec(v)))^(2))/(2)+cdots:}\begin{equation*} =1+\frac{(\vec{u}-\vec{v})^{2}}{2}+\cdots \tag{E.4} \end{equation*}(E.4)=1+(uv)22+
and the two results follow after taking the square root. This is just the Galilean relative velocity, and the two results are found because we have not specified whether the relative velocity is u u uuu relative to v v vvv or the other way around.
(2.3) The first answers follow by simple matrix multiplication using η μ ν η μ ν eta_(mu nu)\eta_{\mu \nu}ημν given in eqn 2.15. Some of the later answers are given in the main text of the chapter.
(2.4) The proton has γ = 10 19 / 10 9 = 10 10 γ = 10 19 / 10 9 = 10 10 gamma=10^(19)//10^(9)=10^(10)\gamma=10^{19} / 10^{9}=10^{10}γ=1019/109=1010 and so is travelling very close to the speed of light. It travels 10 5 10 5 10^(5)10^{5}105 light-years in 10 5 10 5 10^(5)10^{5}105 years. In its rest frame, the galaxy is contracted by a factor γ γ gamma\gammaγ and so the journey is only 10 5 10 5 10^(-5)10^{-5}105 light-years long, taking 10 5 10 5 10^(-5)10^{-5}105 years, or about 5 minutes.
(2.5) The Lorentz transformation gives p μ = ( γ m , γ m v , 0 , 0 ) p μ = ( γ m , γ m v , 0 , 0 ) p^(mu)=(gamma m,gamma mv,0,0)p^{\mu}=(\gamma m, \gamma m v, 0,0)pμ=(γm,γmv,0,0) and so one can just read off E = γ m E = γ m E=gamma mE=\gamma mE=γm (i.e. E = γ m c 2 E = γ m c 2 E=gamma mc^(2)E=\gamma m c^{2}E=γmc2 if you put the factors of c c ccc back in) and p x = γ m v p x = γ m v p^(x)=gamma mvp^{x}=\gamma m vpx=γmv. Both sets of components give p p = ( p 0 ) 2 + ( p 1 ) 2 + ( p 2 ) 2 + p p = p 0 2 + p 1 2 + p 2 2 + p*p=-(p^(0))^(2)+(p^(1))^(2)+(p^(2))^(2)+\boldsymbol{p} \cdot \boldsymbol{p}=-\left(p^{0}\right)^{2}+\left(p^{1}\right)^{2}+\left(p^{2}\right)^{2}+pp=(p0)2+(p1)2+(p2)2+ ( p 3 ) 2 = m 2 p 3 2 = m 2 (p^(3))^(2)=-m^(2)\left(p^{3}\right)^{2}=-m^{2}(p3)2=m2.
(2.6) Assume that we can have e + γ e e + γ e e+gamma rarre\mathrm{e}+\gamma \rightarrow \mathrm{e}e+γe, so that
(E.5) p e + p γ = p e . (E.5) p e + p γ = p e . {:(E.5)p_(e)+p_(gamma)=p_(e)^(').:}\begin{equation*} \boldsymbol{p}_{\mathrm{e}}+\boldsymbol{p}_{\gamma}=\boldsymbol{p}_{\mathrm{e}}^{\prime} . \tag{E.5} \end{equation*}(E.5)pe+pγ=pe.
If we square both sides, we get
p e p e + 2 p e p γ + p γ p γ = p e p e p e p e + 2 p e p γ + p γ p γ = p e p e p_(e)*p_(e)+2p_(e)*p_(gamma)+p_(gamma)*p_(gamma)=p_(e)^(')*p_(e)^(')\boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\mathrm{e}}+2 \boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\gamma}+\boldsymbol{p}_{\gamma} \cdot \boldsymbol{p}_{\gamma}=\boldsymbol{p}_{\mathrm{e}}^{\prime} \cdot \boldsymbol{p}_{\mathrm{e}}^{\prime}pepe+2pepγ+pγpγ=pepe,
but p e p e = p e p e = m e 2 p e p e = p e p e = m e 2 p_(e)*p_(e)=p_(e)^(')*p_(e)^(')=-m_(e)^(2)\boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\mathrm{e}}=\boldsymbol{p}_{\mathrm{e}}^{\prime} \cdot \boldsymbol{p}_{\mathrm{e}}^{\prime}=-m_{\mathrm{e}}^{2}pepe=pepe=me2 and p γ p γ = 0 p γ p γ = 0 p_(gamma)*p_(gamma)=0\boldsymbol{p}_{\gamma} \cdot \boldsymbol{p}_{\gamma}=0pγpγ=0, so we are left with 2 p e p γ = 0 2 p e p γ = 0 2p_(e)*p_(gamma)=02 \boldsymbol{p}_{\mathrm{e}} \cdot \boldsymbol{p}_{\gamma}=02pepγ=0. We can evaluate this in the rest frame of the electron where p e = ( m e , 0 ) p e = m e , 0 p_(e)=(m_(e), vec(0))\boldsymbol{p}_{\mathrm{e}}=\left(m_{\mathrm{e}}, \overrightarrow{0}\right)pe=(me,0) and in general the photon has p γ = ( E , p ) p γ = ( E , p ) p_(gamma)=(E, vec(p))\boldsymbol{p}_{\gamma}=(E, \vec{p})pγ=(E,p) with E = | p | E = | p | E=| vec(p)|E=|\vec{p}|E=|p|. So we are left with E m e = 0 E m e = 0 Em_(e)=0E m_{\mathrm{e}}=0Eme=0, which means that E = 0 E = 0 E=0E=0E=0, i.e. only holds for (non-existent) zero-energy photons.
(2.7) Here the coordinate q q qqq of interest is q = ϕ q = ϕ q=phiq=\phiq=ϕ, the amplitude of the string's displacement. However, we're not interested in ϕ ϕ phi\phiϕ as a function of time, but rather of length x x xxx along the string. This is no problem, we simply proceed with ϕ ˙ = ϕ / x ϕ ˙ = ϕ / x phi^(˙)=del phi//del x\dot{\phi}=\partial \phi / \partial xϕ˙=ϕ/x. Instead of the action, the functional we want to minimize is the energy E [ ϕ ] E [ ϕ ] E[phi]E[\phi]E[ϕ]. There are two contributions to the energy, the strain d ϕ / d x d ϕ / d x propdphi//dx\propto \mathrm{d} \phi / \mathrm{d} xdϕ/dx tells us it costs energy for the string to bend, the gravitational potential energy V = ρ g ϕ ( x ) V = ρ g ϕ ( x ) V=-rho g phi(x)V=-\rho g \phi(x)V=ρgϕ(x) shows that the energy is reduced the large x x xxx. The energy functional is
(E.7) E ( ϕ ) = d x [ T 2 ( d ϕ ( x ) d x ) 2 ρ g ϕ ( x ) ] (E.7) E ( ϕ ) = d x T 2 d ϕ ( x ) d x 2 ρ g ϕ ( x ) {:(E.7)E(phi)=intdx[(T)/(2)(((d)phi(x))/(dx))^(2)-rho g phi(x)]:}\begin{equation*} E(\phi)=\int \mathrm{d} x\left[\frac{T}{2}\left(\frac{\mathrm{~d} \phi(x)}{\mathrm{d} x}\right)^{2}-\rho g \phi(x)\right] \tag{E.7} \end{equation*}(E.7)E(ϕ)=dx[T2( dϕ(x)dx)2ρgϕ(x)]
Notice that this is in the same form as the mechanical action, with the first term playing the role of the kinetic energy. Feeding this through the Euler Lagrange equations, we find
E ( d ϕ d x ) = T d ϕ d x (E.8) d d x E ( d ϕ d x ) = T d 2 ϕ d x 2 E ϕ = ρ g E d ϕ d x = T d ϕ d x (E.8) d d x E d ϕ d x = T d 2 ϕ d x 2 E ϕ = ρ g {:[(del E)/(del(((d)phi)/((d)x)))=T((d)phi)/((d)x)],[(E.8)((d))/((d)x)(del E)/(del(((d)phi)/((d)x)))=T(d^(2)phi)/((d)x^(2))],[(del E)/(del phi)=-rho g]:}\begin{gather*} \frac{\partial E}{\partial\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} x}\right)}=T \frac{\mathrm{~d} \phi}{\mathrm{~d} x} \\ \frac{\mathrm{~d}}{\mathrm{~d} x} \frac{\partial E}{\partial\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} x}\right)}=T \frac{\mathrm{~d}^{2} \phi}{\mathrm{~d} x^{2}} \tag{E.8}\\ \frac{\partial E}{\partial \phi}=-\rho g \end{gather*}E( dϕ dx)=T dϕ dx(E.8) d dxE( dϕ dx)=T d2ϕ dx2Eϕ=ρg
and
on of motion
(E.10) d 2 ϕ = ρ g (E.10) d 2 ϕ _ = ρ g _ {:(E.10)d^(2)phi_=-rho g_:}\begin{equation*} \underline{\mathrm{d}^{2} \phi}=-\underline{\rho g} \tag{E.10} \end{equation*}(E.10)d2ϕ=ρg
The solution tells us the shape of the hanging string, which is a parabola. This can be written as
(E.11) ϕ ( x ) = ρ g 2 T [ ( L 2 ) 2 x 2 ] (E.11) ϕ ( x ) = ρ g 2 T L 2 2 x 2 {:(E.11)phi(x)=(rho g)/(2T)[((L)/(2))^(2)-x^(2)]:}\begin{equation*} \phi(x)=\frac{\rho g}{2 T}\left[\left(\frac{L}{2}\right)^{2}-x^{2}\right] \tag{E.11} \end{equation*}(E.11)ϕ(x)=ρg2T[(L2)2x2]
(2.8) We have that p i = p i = L v i = v i ( m 1 v j v j ) = p i = p i = L v i = v i m 1 v j v j = p^(i)=p_(i)=(del L)/(delv^(i))=(del)/(delv^(i))(-msqrt(1-v^(j)v^(j)))=p^{i}=p_{i}=\frac{\partial L}{\partial v^{i}}=\frac{\partial}{\partial v^{i}}\left(-m \sqrt{1-v^{j} v^{j}}\right)=pi=pi=Lvi=vi(m1vjvj)= m 2 2 v i 1 v j v j m 2 2 v i 1 v j v j -(m)/(2)*(-2v^(i))/(sqrt(1-v^(j)v^(j)))-\frac{m}{2} \cdot \frac{-2 v^{i}}{\sqrt{1-v^{j} v^{j}}}m22vi1vjvj, and hence p i = m γ v i p i = m γ v i p^(i)=m gammav^(i)p^{i}=m \gamma v^{i}pi=mγvi.
(2.9) H = p v L = m γ v 2 + m γ = m γ ( v 2 + 1 v 2 ) = γ m H = p v L = m γ v 2 + m γ = m γ v 2 + 1 v 2 = γ m H= vec(p)* vec(v)-L=m gammav^(2)+(m)/( gamma)=m gamma(v^(2)+1-v^(2))=gamma mH=\vec{p} \cdot \vec{v}-L=m \gamma v^{2}+\frac{m}{\gamma}=m \gamma\left(v^{2}+1-v^{2}\right)=\gamma mH=pvL=mγv2+mγ=mγ(v2+1v2)=γm.
(3.2) We write
e r θ = θ ( cos θ e x + sin θ e y ) (E.12) = sin θ e x + cos θ e y = e θ r e r θ = θ cos θ e x + sin θ e y (E.12) = sin θ e x + cos θ e y = e θ r {:[(dele_(r))/(del theta)=(del)/(del theta)(cos thetae_(x)+sin thetae_(y))],[(E.12)=-sin thetae_(x)+cos thetae_(y)=(e_(theta))/(r)]:}\begin{align*} \frac{\partial \boldsymbol{e}_{r}}{\partial \theta} & =\frac{\partial}{\partial \theta}\left(\cos \theta \boldsymbol{e}_{x}+\sin \theta \boldsymbol{e}_{y}\right) \\ & =-\sin \theta \boldsymbol{e}_{x}+\cos \theta \boldsymbol{e}_{y}=\frac{\boldsymbol{e}_{\theta}}{r} \tag{E.12} \end{align*}erθ=θ(cosθex+sinθey)(E.12)=sinθex+cosθey=eθr
We also work out that
e θ θ = θ ( r sin θ e x + r cos θ e y ) (E.13) = r cos θ e x r sin θ e y = r e r e θ θ = θ r sin θ e x + r cos θ e y (E.13) = r cos θ e x r sin θ e y = r e r {:[(dele_(theta))/(del theta)=(del)/(del theta)(-r sin thetae_(x)+r cos thetae_(y))],[(E.13)=-r cos thetae_(x)-r sin thetae_(y)=-re_(r)]:}\begin{align*} \frac{\partial \boldsymbol{e}_{\theta}}{\partial \theta} & =\frac{\partial}{\partial \theta}\left(-r \sin \theta \boldsymbol{e}_{x}+r \cos \theta \boldsymbol{e}_{y}\right) \\ & =-r \cos \theta \boldsymbol{e}_{x}-r \sin \theta \boldsymbol{e}_{y}=-r \boldsymbol{e}_{r} \tag{E.13} \end{align*}eθθ=θ(rsinθex+rcosθey)(E.13)=rcosθexrsinθey=rer
The final two results follow from taking derivatives with respect to r r rrr
(E.14) e r r = 0 , e θ r = e θ r . (E.14) e r r = 0 , e θ r = e θ r . {:(E.14)(dele_(r))/(del r)=0","quad(dele_(theta))/(del r)=(e_(theta))/(r).:}\begin{equation*} \frac{\partial \boldsymbol{e}_{r}}{\partial r}=0, \quad \frac{\partial \boldsymbol{e}_{\theta}}{\partial r}=\frac{\boldsymbol{e}_{\theta}}{r} . \tag{E.14} \end{equation*}(E.14)err=0,eθr=eθr.
(3.3) Expanding sin ( r / K ) sin ( r / K ) sin(r//sqrtK)\sin (r / \sqrt{K})sin(r/K) for r K r K r≪sqrtKr \ll \sqrt{K}rK gives C = 2 π r = C = 2 π r = C=2pi r=C=2 \pi r=C=2πr= ( 1 K r 2 6 + ) 1 K r 2 6 + (1-(Kr^(2))/(6)+dots)\left(1-\frac{K r^{2}}{6}+\ldots\right)(1Kr26+) and rearranging gives the desired result.
(3.5) Substituting we obtain
d s 2 = d t 2 [ 1 Ω 2 ( x 2 + y 2 ) ] + d x 2 + d y 2 + d z 2 2 Ω y d x d t + 2 Ω x d y d t . d s 2 = d t 2 1 Ω 2 x 2 + y 2 + d x 2 + d y 2 + d z 2 2 Ω y d x d t + 2 Ω x d y d t . {:[ds^(2)=-dt^('2)[1-Omega^(2)(x^('2)+y^('2))]+dx^('2)+dy^('2)],[+dz^('2)-2Omegay^(')dx^(')dt^(')+2Omegax^(')dy^(')dt^(').]:}\begin{aligned} \mathrm{d} s^{2}= & -\mathrm{d} t^{\prime 2}\left[1-\Omega^{2}\left(x^{\prime 2}+y^{\prime 2}\right)\right]+\mathrm{d} x^{\prime 2}+\mathrm{d} y^{\prime 2} \\ & +\mathrm{d} z^{\prime 2}-2 \Omega y^{\prime} \mathrm{d} x^{\prime} \mathrm{d} t^{\prime}+2 \Omega x^{\prime} \mathrm{d} y^{\prime} \mathrm{d} t^{\prime} . \end{aligned}ds2=dt2[1Ω2(x2+y2)]+dx2+dy2+dz22Ωydxdt+2Ωxdydt.
(4.1) Since u u u\boldsymbol{u}u and v v v\boldsymbol{v}v transform as 4 -vectors, the tensor transformation law gives Λ α α Λ β β W α β = Λ α α β β u α v β = Λ α α Λ β β W α β = Λ α α β β u α v β = Lambda^(alpha^('))_(alpha)Lambda^(beta^('))_(beta)W^(alpha beta)=Lambda^(alpha^('))_(alpha)^(beta^('))_(beta)u^(alpha)v^(beta)=\Lambda^{\alpha^{\prime}}{ }_{\alpha} \Lambda^{\beta^{\prime}}{ }_{\beta} W^{\alpha \beta}=\Lambda^{\alpha^{\prime}}{ }_{\alpha}{ }^{\beta^{\prime}}{ }_{\beta} u^{\alpha} v^{\beta}=ΛααΛββWαβ=Λααββuαvβ= u α v β = W α β u α v β = W α β u^(alpha^('))v^(beta^('))=W^(alpha^(')beta^('))u^{\alpha^{\prime}} v^{\beta^{\prime}}=W^{\alpha^{\prime} \beta^{\prime}}uαvβ=Wαβ as required.
(4.2) Λ α α Λ β β δ α β = Λ α β Λ β β = δ α Λ α α Λ β β δ α β = Λ α β Λ β β = δ α Lambda^(alpha^('))_(alpha)Lambda^(beta)_(beta^('))delta^(alpha)_(beta)=Lambda^(alpha^('))_(beta)Lambda^(beta)_(beta^('))=delta^(alpha^('))_(^('))\Lambda^{\alpha^{\prime}}{ }_{\alpha} \Lambda^{\beta}{ }_{\beta^{\prime}} \delta^{\alpha}{ }_{\beta}=\Lambda^{\alpha^{\prime}}{ }_{\beta} \Lambda^{\beta}{ }_{\beta^{\prime}}=\delta^{\alpha^{\prime}}{ }_{{ }^{\prime}}ΛααΛββδαβ=ΛαβΛββ=δα as required for it to transform properly. We can use the metric tensor to raise and lower the indices in δ α β δ α β delta^(alpha beta)\delta^{\alpha \beta}δαβ and δ α β δ α β delta_(alpha beta)\delta_{\alpha \beta}δαβ as appropriate to turn them into δ α β δ α β delta^(alpha)_(beta)\delta^{\alpha}{ }_{\beta}δαβ (which we've just shown works as a tensor)
δ α β = η α γ δ β γ = η α β , (E.16) δ α β = η α γ δ γ β = η α β . δ α β = η α γ δ β γ = η α β , (E.16) δ α β = η α γ δ γ β = η α β . {:[delta^(alpha beta)=eta^(alpha gamma)delta^(beta)_(gamma)=eta^(alpha beta)","],[(E.16)delta_(alpha beta)=eta_(alpha gamma)delta^(gamma)_(beta)=eta_(alpha beta).]:}\begin{align*} & \delta^{\alpha \beta}=\eta^{\alpha \gamma} \delta^{\beta}{ }_{\gamma}=\eta^{\alpha \beta}, \\ & \delta_{\alpha \beta}=\eta_{\alpha \gamma} \delta^{\gamma}{ }_{\beta}=\eta_{\alpha \beta} . \tag{E.16} \end{align*}δαβ=ηαγδβγ=ηαβ,(E.16)δαβ=ηαγδγβ=ηαβ.
Thus, if you try and write δ α β δ α β delta_(alpha beta)\delta_{\alpha \beta}δαβ and δ α β δ α β delta^(alpha beta)\delta^{\alpha \beta}δαβ as tensors, they end up being equal to the metric tensor and lose their Kronecker-delta property. Thus, it is best to keep δ α β δ α β delta_(alpha beta)\delta_{\alpha \beta}δαβ and δ α β δ α β delta^(alpha beta)\delta^{\alpha \beta}δαβ as non-tensor objects which are shorthands for quantities equalling one when α = β α = β alpha=beta\alpha=\betaα=β and zero otherquantities equalling one when α = β α = β alpha=beta\alpha=\betaα=β if you want the Kronecker delta to behave as a
wise. wise. If you want the Kronecker delta to be
tensor, you have to use the mixed form δ α β δ α β delta^(alpha)_(beta)\delta^{\alpha}{ }_{\beta}δαβ.
(4.3) S S S\boldsymbol{S}S is a tensor and so its components transforms properly
(E.17) S ν μ = Λ μ Λ ν ν S ν μ (E.17) S ν μ = Λ μ Λ ν ν S ν μ {:(E.17)S_(nu^('))^(mu^('))=Lambda^(mu^('))Lambda_(nu^('))^(nu)S_(nu)^(mu):}\begin{equation*} S_{\nu^{\prime}}^{\mu^{\prime}}=\Lambda^{\mu^{\prime}} \Lambda_{\nu^{\prime}}^{\nu} S_{\nu}^{\mu} \tag{E.17} \end{equation*}(E.17)Sνμ=ΛμΛννSνμ
Thus,
(E.18) S μ μ = Λ μ μ Λ μ ν S ν μ = δ ν μ S ν μ = S μ μ . (E.18) S μ μ = Λ μ μ Λ μ ν S ν μ = δ ν μ S ν μ = S μ μ . {:(E.18)S_(mu^('))^(mu^('))=Lambda_(mu)^(mu^('))Lambda_(mu^('))^(nu)S_(nu)^(mu)=delta^(nu)_(mu)S_(nu)^(mu)=S_(mu)^(mu).:}\begin{equation*} S_{\mu^{\prime}}^{\mu^{\prime}}=\Lambda_{\mu}^{\mu^{\prime}} \Lambda_{\mu^{\prime}}^{\nu} S_{\nu}^{\mu}=\delta^{\nu}{ }_{\mu} S_{\nu}^{\mu}=S_{\mu}^{\mu} . \tag{E.18} \end{equation*}(E.18)Sμμ=ΛμμΛμνSνμ=δνμSνμ=Sμμ.
(4.4) (a) Start with the relationships
r = x 2 + y 2 + z 2 , cos θ = z r , (E.19) tan ϕ = y x . r = x 2 + y 2 + z 2 , cos θ = z r , (E.19) tan ϕ = y x . {:[r=x^(2)+y^(2)+z^(2)","],[cos theta=(z)/(r)","],[(E.19)tan phi=(y)/(x).]:}\begin{align*} r & =x^{2}+y^{2}+z^{2}, \\ \cos \theta & =\frac{z}{r}, \\ \tan \phi & =\frac{y}{x} . \tag{E.19} \end{align*}r=x2+y2+z2,cosθ=zr,(E.19)tanϕ=yx.
Using the transformation law
(E.20) ω μ = ( x μ x ν ) ω ν (E.20) ω μ = x μ x ν ω ν {:(E.20)omega^(mu)=((delx^(mu))/(delx^(nu)))omega^(nu):}\begin{equation*} \boldsymbol{\omega}^{\mu}=\left(\frac{\partial x^{\mu}}{\partial x^{\nu}}\right) \boldsymbol{\omega}^{\nu} \tag{E.20} \end{equation*}(E.20)ωμ=(xμxν)ων
we find basis 1-forms
ω r = sin θ cos ϕ ω x + sin θ sin ϕ ω y + cos θ ω z ω θ = cos θ cos ϕ r ω x + cos θ sin ϕ r ω y sin θ r ω z (E.21) ω ϕ = sin ϕ r sin θ ω x + cos ϕ r sin θ ω y ω r = sin θ cos ϕ ω x + sin θ sin ϕ ω y + cos θ ω z ω θ = cos θ cos ϕ r ω x + cos θ sin ϕ r ω y sin θ r ω z (E.21) ω ϕ = sin ϕ r sin θ ω x + cos ϕ r sin θ ω y {:[omega^(r)=sin theta cos phiomega^(x)+sin theta sin phiomega^(y)+cos thetaomega^(z)],[omega^(theta)=(cos theta cos phi)/(r)omega^(x)+(cos theta sin phi)/(r)omega^(y)-(sin theta)/(r)omega^(z)],[(E.21)omega^(phi)=-(sin phi)/(r sin theta)omega^(x)+(cos phi)/(r sin theta)omega^(y)]:}\begin{align*} \boldsymbol{\omega}^{r} & =\sin \theta \cos \phi \boldsymbol{\omega}^{x}+\sin \theta \sin \phi \boldsymbol{\omega}^{y}+\cos \theta \boldsymbol{\omega}^{z} \\ \boldsymbol{\omega}^{\theta} & =\frac{\cos \theta \cos \phi}{r} \boldsymbol{\omega}^{x}+\frac{\cos \theta \sin \phi}{r} \boldsymbol{\omega}^{y}-\frac{\sin \theta}{r} \boldsymbol{\omega}^{z} \\ \boldsymbol{\omega}^{\phi} & =-\frac{\sin \phi}{r \sin \theta} \boldsymbol{\omega}^{x}+\frac{\cos \phi}{r \sin \theta} \boldsymbol{\omega}^{y} \tag{E.21} \end{align*}ωr=sinθcosϕωx+sinθsinϕωy+cosθωzωθ=cosθcosϕrωx+cosθsinϕrωysinθrωz(E.21)ωϕ=sinϕrsinθωx+cosϕrsinθωy
(b) This time we start with
x = r sin θ cos ϕ , y = r sin θ sin ϕ , (E.22) z = r cos θ x = r sin θ cos ϕ , y = r sin θ sin ϕ , (E.22) z = r cos θ {:[x=r sin theta cos phi","],[y=r sin theta sin phi","],[(E.22)z=r cos theta]:}\begin{align*} & x=r \sin \theta \cos \phi, \\ & y=r \sin \theta \sin \phi, \\ & z=r \cos \theta \tag{E.22} \end{align*}x=rsinθcosϕ,y=rsinθsinϕ,(E.22)z=rcosθ
Using the transformation law for basis vectors e μ = e μ = e_(mu)=e_{\mu}=eμ= ( x ν / x μ ) e ν x ν / x μ e ν (delx^(nu)//delx^(mu))e_(nu)\left(\partial x^{\nu} / \partial x^{\mu}\right) \boldsymbol{e}_{\nu}(xν/xμ)eν, we find that the basis vectors are
e r = sin θ cos ϕ e x + sin θ sin ϕ e y + cos θ e z e θ = cos θ cos ϕ e x + cos θ sin ϕ e y sin θ e z e ϕ = r sin θ sin ϕ e x + r sin θ cos ϕ e y e r = sin θ cos ϕ e x + sin θ sin ϕ e y + cos θ e z e θ = cos θ cos ϕ e x + cos θ sin ϕ e y sin θ e z e ϕ = r sin θ sin ϕ e x + r sin θ cos ϕ e y {:[e_(r)=sin theta cos phie_(x)+sin theta sin phie_(y)+cos thetae_(z)],[e_(theta)=cos theta cos phie_(x)+cos theta sin phie_(y)-sin thetae_(z)],[e_(phi)=-r sin theta sin phie_(x)+r sin theta cos phie_(y)]:}\begin{aligned} & \boldsymbol{e}_{r}=\sin \theta \cos \phi \boldsymbol{e}_{x}+\sin \theta \sin \phi \boldsymbol{e}_{y}+\cos \theta \boldsymbol{e}_{z} \\ & \boldsymbol{e}_{\theta}=\cos \theta \cos \phi \boldsymbol{e}_{x}+\cos \theta \sin \phi \boldsymbol{e}_{y}-\sin \theta \boldsymbol{e}_{z} \\ & \boldsymbol{e}_{\phi}=-r \sin \theta \sin \phi \boldsymbol{e}_{x}+r \sin \theta \cos \phi \boldsymbol{e}_{y} \end{aligned}er=sinθcosϕex+sinθsinϕey+cosθezeθ=cosθcosϕex+cosθsinϕeysinθezeϕ=rsinθsinϕex+rsinθcosϕey
(c) Notice that the components for the vector e r e r e_(r)\boldsymbol{e}_{r}er and 1 -form ω r ω r omega^(r)\boldsymbol{\omega}^{r}ωr are the same; those for θ θ theta\thetaθ related by a factor r r rrr and those for ϕ ϕ phi\phiϕ by a factor r 2 sin 2 θ r 2 sin 2 θ r^(2)sin^(2)thetar^{2} \sin ^{2} \thetar2sin2θ. The orthogonality condition ω μ , e ν = δ μ ν ω μ , e ν = δ μ ν (:omega^(mu),e_(nu):)=delta^(mu)_(nu)\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\nu}\right\rangle=\delta^{\mu}{ }_{\nu}ωμ,eν=δμν is then guaranteed by the orthogonality of the basis vectors e ν e ν e_(nu)\boldsymbol{e}_{\nu}eν (which can be checked).
(4.5) (a) Using e u = ( x μ / u ) e μ e u = x μ / u e μ e_(u)=(delx^(mu)//del u)e_(mu)\boldsymbol{e}_{u}=\left(\partial x^{\mu} / \partial u\right) \boldsymbol{e}_{\mu}eu=(xμ/u)eμ, and so on, we find
e u = e x + e y 2 v e z , e v = e x + e y 2 u e z , (E.24) e w = e z e u = e x + e y 2 v e z , e v = e x + e y 2 u e z , (E.24) e w = e z {:[e_(u)=e_(x)+e_(y)-2ve_(z)","],[e_(v)=-e_(x)+e_(y)-2ue_(z)","],[(E.24)e_(w)=e_(z)]:}\begin{align*} \boldsymbol{e}_{u} & =\boldsymbol{e}_{x}+\boldsymbol{e}_{y}-2 v \boldsymbol{e}_{z}, \\ \boldsymbol{e}_{v} & =-\boldsymbol{e}_{x}+\boldsymbol{e}_{y}-2 u \boldsymbol{e}_{z}, \\ \boldsymbol{e}_{w} & =\boldsymbol{e}_{z} \tag{E.24} \end{align*}eu=ex+ey2vez,ev=ex+ey2uez,(E.24)ew=ez
(b) Using ω u = ( u / x μ ) ω μ ω u = u / x μ ω μ omega^(u)=(del u//delx^(mu))omega^(mu)\boldsymbol{\omega}^{u}=\left(\partial u / \partial x^{\mu}\right) \boldsymbol{\omega}^{\mu}ωu=(u/xμ)ωμ and so on, we have
ω u = 1 2 ω x + 1 2 ω y ω v = 1 2 ω x + 1 2 ω y ω u = 1 2 ω x + 1 2 ω y ω v = 1 2 ω x + 1 2 ω y {:[omega^(u)=(1)/(2)omega^(x)+(1)/(2)omega^(y)],[omega^(v)=-(1)/(2)*omega^(x)+(1)/(2)*omega^(y)]:}\begin{aligned} \boldsymbol{\omega}^{u} & =\frac{1}{2} \boldsymbol{\omega}^{x}+\frac{1}{2} \boldsymbol{\omega}^{y} \\ \boldsymbol{\omega}^{v} & =-\frac{1}{2} \cdot \boldsymbol{\omega}^{x}+\frac{1}{2} \cdot \boldsymbol{\omega}^{y} \end{aligned}ωu=12ωx+12ωyωv=12ωx+12ωy
(E.25) ω w = x ω x + y ω y + ω z (E.25) ω w = x ω x + y ω y + ω z {:(E.25)omega^(w)=-xomega^(x)+yomega^(y)+omega^(z):}\begin{equation*} \boldsymbol{\omega}^{w}=-x \boldsymbol{\omega}^{x}+y \boldsymbol{\omega}^{y}+\boldsymbol{\omega}^{z} \tag{E.25} \end{equation*}(E.25)ωw=xωx+yωy+ωz
Note that the system is not orthogonal.
(4.8) The transformation law is
(E.26) ( ω k x k y k z ) = ( γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 ) ( ω k x k y k z ) (E.26) ω k x k y k z = γ β γ 0 0 β γ γ 0 0 0 0 1 0 0 0 0 1 ω k x k y k z {:(E.26)([omega^(')],[k^(x^('))],[k^(y^('))],[k^(z^('))])=([gamma,-beta gamma,0,0],[-beta gamma,gamma,0,0],[0,0,1,0],[0,0,0,1])([omega],[k^(x)],[k^(y)],[k^(z)]):}\left(\begin{array}{l} \omega^{\prime} \tag{E.26}\\ k^{x^{\prime}} \\ k^{y^{\prime}} \\ k^{z^{\prime}} \end{array}\right)=\left(\begin{array}{cccc} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c} \omega \\ k^{x} \\ k^{y} \\ k^{z} \end{array}\right)(E.26)(ωkxkykz)=(γβγ00βγγ0000100001)(ωkxkykz)
We obtain
(E.27) ω = γ ( ω β k x ) (E.27) ω = γ ω β k x {:(E.27)omega^(')=gamma(omega-betak^(x)):}\begin{equation*} \omega^{\prime}=\gamma\left(\omega-\beta k^{x}\right) \tag{E.27} \end{equation*}(E.27)ω=γ(ωβkx)
Setting k x = | k | cos α = ω cos α k x = | k | cos α = ω cos α k^(x)=| vec(k)|cos alpha=omega cos alphak^{x}=|\vec{k}| \cos \alpha=\omega \cos \alphakx=|k|cosα=ωcosα for a light source whose direction of emission makes an angle α α alpha\alphaα with the x x xxx direction, we have
(E.28) ω = γ ω ( 1 β cos α ) (E.28) ω = γ ω ( 1 β cos α ) {:(E.28)omega^(')=gamma omega(1-beta cos alpha):}\begin{equation*} \omega^{\prime}=\gamma \omega(1-\beta \cos \alpha) \tag{E.28} \end{equation*}(E.28)ω=γω(1βcosα)
If the source is in the primed frame where it is measured to have a frequency ω 0 ω 0 omega_(0)\omega_{0}ω0, we have
(E.29) ω ω 0 = ( 1 β 2 ) 1 2 1 β cos α (E.29) ω ω 0 = 1 β 2 1 2 1 β cos α {:(E.29)(omega)/(omega_(0))=((1-beta^(2))^((1)/(2)))/(1-beta cos alpha):}\begin{equation*} \frac{\omega}{\omega_{0}}=\frac{\left(1-\beta^{2}\right)^{\frac{1}{2}}}{1-\beta \cos \alpha} \tag{E.29} \end{equation*}(E.29)ωω0=(1β2)121βcosα
(5.3) (b) Dropping primes, setting d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 and dividing through by d t 2 d t 2 dt^(2)\mathrm{d} t^{2}dt2 we find, in the x t x t x-tx-txt plane, that
(E.30) 1 v 2 = ( d x d t ) 2 2 v d x d t (E.30) 1 v 2 = d x d t 2 2 v d x d t {:(E.30)1-v^(2)=((dx)/((d)t))^(2)-2v((d)x)/((d)t):}\begin{equation*} 1-v^{2}=\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}-2 v \frac{\mathrm{~d} x}{\mathrm{~d} t} \tag{E.30} \end{equation*}(E.30)1v2=(dx dt)22v dx dt
Solving this quadratic equation, we obtain an equation for the light cones of
(E.31) ( d x d t ) = v ± 1 (E.31) d x d t = v ± 1 {:(E.31)((dx)/((d)t))=v+-1:}\begin{equation*} \left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)=v \pm 1 \tag{E.31} \end{equation*}(E.31)(dx dt)=v±1
which tells us that the light cones make an angle tan 1 ( v ± 1 ) tan 1 ( v ± 1 ) tan^(-1)(v+-1)\tan ^{-1}(v \pm 1)tan1(v±1) to the t t ttt axis.
(c) We obtain the usual expression for time dilation in
flat spacetime d τ = d t ( 1 v 2 ) 1 2 d τ = d t 1 v 2 1 2 dtau=dt(1-v^(2))^((1)/(2))\mathrm{d} \tau=\mathrm{d} t\left(1-v^{2}\right)^{\frac{1}{2}}dτ=dt(1v2)12.
(d) We find
(E.32) g μ ν = ( 1 v 0 0 v 1 v 2 0 0 0 0 1 0 0 0 0 1 ) (E.32) g μ ν = 1 v 0 0 v 1 v 2 0 0 0 0 1 0 0 0 0 1 {:(E.32)g^(mu nu)=([-1,-v,0,0],[-v,1-v^(2),0,0],[0,0,1,0],[0,0,0,1]):}g^{\mu \nu}=\left(\begin{array}{cccc} -1 & -v & 0 & 0 \tag{E.32}\\ -v & 1-v^{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(E.32)gμν=(1v00v1v20000100001)
(5.4) (a) Setting d t = 0 d t = 0 dt=0\mathrm{d} t=0dt=0, we have a proper length interval
(E.33) d l = a ( t ) d r ( 1 k r 2 ) 1 2 (E.33) d l = a ( t ) d r 1 k r 2 1 2 {:(E.33)dl=(a(t)dr)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \mathrm{d} l=\frac{a(t) \mathrm{d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{E.33} \end{equation*}(E.33)dl=a(t)dr(1kr2)12
(5.5) (a) d l = d x d l = d x dl=dx\mathrm{d} l=\mathrm{d} xdl=dx.
(b) d τ = x d t d τ = x d t dtau=xdt\mathrm{d} \tau=x \mathrm{~d} tdτ=x dt, where x x xxx is the position of the event.
(6.1) The time dilation factor is 1 2 G M / r c 2 1 2 G M / r c 2 sqrt(1-2GM//rc^(2))\sqrt{1-2 G M / r c^{2}}12GM/rc2 and the numerical values come out to be (a) 1 1.1 × 10 8 1 1.1 × 10 8 1-1.1 xx10^(-8)1-1.1 \times 10^{-8}11.1×108 (b) 1 2.1 × 10 6 1 2.1 × 10 6 1-2.1 xx10^(-6)1-2.1 \times 10^{-6}12.1×106 (c) 0.84 . All three values are a bit less than 1, but for (c) it's quite a bit less. Note that for the time dilation factor on the surface of the Earth (comparing it to zero gravitational field) you have to include the effect of the Sun's gravitational field at the Earth's surface, so that the answer is obtained by computing 1 2 G / c 2 ( M / R + M / R ) G / c 2 ( M / R + 1 2 G / c 2 M / R + M / R G / c 2 M / R + sqrt(1-2G//c^(2)(M_(o+)//R_(o+)+M_(o.)//R))~~G//c^(2)(M_(o+)//R_(o+)+:}\sqrt{1-2 G / c^{2}\left(M_{\oplus} / R_{\oplus}+M_{\odot} / R\right)} \approx G / c^{2}\left(M_{\oplus} / R_{\oplus}+\right.12G/c2(M/R+M/R)G/c2(M/R+ M / R M / R M_(o.)//RM_{\odot} / RM/R ), where R R RRR is an astronomical unit.
(6.2) Differentiating the 1 2 G M / r c 2 1 2 G M / r c 2 sqrt(1-2GM_(o+)//rc^(2))\sqrt{1-2 G M_{\oplus} / r c^{2}}12GM/rc2 factor gives G M / r 2 c 2 = g / c 2 = 1.1 × 10 16 m 1 = 1.1 × G M / r 2 c 2 = g / c 2 = 1.1 × 10 16 m 1 = 1.1 × -GM_(o+)//r^(2)c^(2)=-g//c^(2)=-1.1 xx10^(-16)m^(-1)=-1.1 xx-G M_{\oplus} / r^{2} c^{2}=-g / c^{2}=-1.1 \times 10^{-16} \mathrm{~m}^{-1}=-1.1 \timesGM/r2c2=g/c2=1.1×1016 m1=1.1× 10 19 mm 1 10 19 mm 1 10^(-19)mm^(-1)10^{-19} \mathrm{~mm}^{-1}1019 mm1, which is consistent with the value given. (Note that, in comparison with the previous exercise, the effect of the Sun is not needed. Although the Sun's gravitational field leads to a substantial time dilation on the Earth's surface, its gradient is much smaller and can be neglected.)
(6.4) (a) Δ τ = ( 1 2 M r ) 1 2 d t Δ τ = 1 2 M r 1 2 d t Delta tau=(1-(2M)/(r))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} tΔτ=(12Mr)12 dt.
(b) Δ τ = ( 1 2 M r 2 ) 1 2 d t Δ τ = 1 2 M r 2 1 2 d t Delta tau=(1-(2M)/(r_(2)))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r_{2}}\right)^{\frac{1}{2}} \mathrm{~d} tΔτ=(12Mr2)12 dt, where d t d t dt\mathrm{d} tdt is the coordinate time interval between wavefront events.
(c) Since the observers are parallel, the interval between the wavefronts being received is the same as the interval between which they are emitted. As a result, the observer at r 2 r 2 r_(2)r_{2}r2 measures an interval Δ τ = ( 1 2 M r 1 ) 1 2 d t Δ τ = 1 2 M r 1 1 2 d t Delta tau=(1-(2M)/(r_(1)))^((1)/(2))dt\Delta \tau=\left(1-\frac{2 M}{r_{1}}\right)^{\frac{1}{2}} \mathrm{~d} tΔτ=(12Mr1)12 dt.
(6.6) (a) Figure 6.6 shows the set up. The time on A's world line will be
(E.34) x 0 + d x 1 0 + d x 2 0 2 (E.34) x 0 + d x 1 0 + d x 2 0 2 {:(E.34)x^(0)+(dx_(1)^(0)+dx_(2)^(0))/(2):}\begin{equation*} x^{0}+\frac{\mathrm{d} x_{1}^{0}+\mathrm{d} x_{2}^{0}}{2} \tag{E.34} \end{equation*}(E.34)x0+dx10+dx202
The two intervals are the two solutions in eqn 6.15, whose sum gives the quoted integrand.
(c) The optical path length is 2 π r + 2 π Ω r 2 2 π r + 2 π Ω r 2 2pi r+2pi Omegar^(2)2 \pi r+2 \pi \Omega r^{2}2πr+2πΩr2. For a counter-propagating beam, the optical path is 2 π r 2 π r 2pi r-2 \pi r-2πr 2 π Ω r 2 2 π Ω r 2 2pi Omegar^(2)2 \pi \Omega r^{2}2πΩr2. The fringe shift is the difference in optical path, divided by the wavelength of the light.
7.1) (i) u v v u u v v u grad_(u)v-grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}uvvu.
(ii) v w ~ ( u ) v w ~ ( u ) grad_(v) tilde(w)(u)\boldsymbol{\nabla}_{\boldsymbol{v}} \tilde{\boldsymbol{w}}(\boldsymbol{u})vw~(u).
(iii) In slot-machine notation, we could rewrite the expression ( u T ) ( v , w ) u T ( v , w ) (grad_(u)T)(v,w)\left(\nabla_{u} \boldsymbol{T}\right)(\boldsymbol{v}, \boldsymbol{w})(uT)(v,w), becoming u μ T α β ; μ v α w β u μ T α β ; μ v α w β u^(mu)T_(alpha beta;mu)v^(alpha)w^(beta)u^{\mu} T_{\alpha \beta ; \mu} v^{\alpha} w^{\beta}uμTαβ;μvαwβ.
(7.2) (a) The components of the tangent are given by u μ = u μ = u^(mu)=u^{\mu}=uμ= ( d x / d s , d y / d s ) ( d x / d s , d y / d s ) (dx//ds,dy//ds)(\mathrm{d} x / \mathrm{d} s, \mathrm{~d} y / \mathrm{d} s)(dx/ds, dy/ds) and so u μ = ( sin s , cos s ) u μ = ( sin s , cos s ) u^(mu)=(-sin s,cos s)u^{\mu}=(-\sin s, \cos s)uμ=(sins,coss). This vector has unit length.
(b) The curve is not a geodesic, so we do not expect the components of the gradient to vanish. Differentiating, we find ( D u / d s ) μ = ( cos s , sin s ) ( D u / d s ) μ = ( cos s , sin s ) (Du//ds)^(mu)=(-cos s,-sin s)(\mathrm{D} \boldsymbol{u} / \mathrm{d} s)^{\mu}=(-\cos s,-\sin s)(Du/ds)μ=(coss,sins), which gives zero
when dotted with u u u\boldsymbol{u}u
(c) The tangent vector to the reparametrized curve has components u μ = ( t / 1 t 2 , 1 ) u μ = t / 1 t 2 , 1 u^(mu)=(-t//sqrt(1-t^(2)),1)u^{\mu}=\left(-t / \sqrt{1-t^{2}}, 1\right)uμ=(t/1t2,1), from which we find u u = ( 1 t 2 ) 1 u u = 1 t 2 1 u*u=(1-t^(2))^(-1)\boldsymbol{u} \cdot \boldsymbol{u}=\left(1-t^{2}\right)^{-1}uu=(1t2)1. The gradient of this vector has components ( D u / d t ) μ = ( ( 1 t 2 ) 3 2 , 0 ) ( D u / d t ) μ = 1 t 2 3 2 , 0 (Du//dt)^(mu)=(-(1-t^(2))^(-(3)/(2)),0)(\mathrm{D} \boldsymbol{u} / \mathrm{d} t)^{\mu}=\left(-\left(1-t^{2}\right)^{-\frac{3}{2}}, 0\right)(Du/dt)μ=((1t2)32,0) and so u D u / d t = t / ( 1 t 2 ) 2 u D u / d t = t / 1 t 2 2 u*Du//dt=t//(1-t^(2))^(2)\boldsymbol{u} \cdot \mathrm{D} \boldsymbol{u} / \mathrm{d} t=t /\left(1-t^{2}\right)^{2}uDu/dt=t/(1t2)2
The fact that the tangent vector does not have a constant magnitude makes this parametrization less convenient than the (affine) length parametrization that uses s s sss.
(7.3) (a) We find
(E.35) x v = ( 0 C ) , y v = ( 0 0 ) (E.35) x v = ( 0 C ) , y v = ( 0 0 ) {:(E.35)grad_(x)v=((0)/(C))","quadgrad_(y)v=((0)/(0)):}\begin{equation*} \nabla_{x} v=\binom{0}{C}, \quad \nabla_{y} v=\binom{0}{0} \tag{E.35} \end{equation*}(E.35)xv=(0C),yv=(00)
(b) In cylindrical polars, we have
(E.36) ( v r , v θ ) = ( C r sin θ cos θ , C cos 2 θ ) (E.36) v r , v θ = C r sin θ cos θ , C cos 2 θ {:(E.36)(v^(r),v^(theta))=(Cr sin theta cos theta,Ccos^(2)theta):}\begin{equation*} \left(v^{r}, v^{\theta}\right)=\left(C r \sin \theta \cos \theta, C \cos ^{2} \theta\right) \tag{E.36} \end{equation*}(E.36)(vr,vθ)=(Crsinθcosθ,Ccos2θ)
(c) The derivatives are
and
(E.37) r v = ( C sin θ cos θ C r cos 2 θ ) , (E.38) θ v = ( C r sin 2 θ C cos θ sin θ ) . (E.37) r v = ( C sin θ cos θ C r cos 2 θ ) , (E.38) θ v = ( C r sin 2 θ C cos θ sin θ ) . {:[(E.37)grad_(r)v=((C sin theta cos theta)/((C)/(r)*cos^(2)theta))","],[(E.38)grad_(theta)v=((-Crsin^(2)theta)/(-C cos theta sin theta)).]:}\begin{gather*} \boldsymbol{\nabla}_{r} \boldsymbol{v}=\binom{C \sin \theta \cos \theta}{\frac{C}{r} \cdot \cos ^{2} \theta}, \tag{E.37}\\ \boldsymbol{\nabla}_{\theta} \boldsymbol{v}=\binom{-C r \sin ^{2} \theta}{-C \cos \theta \sin \theta} . \tag{E.38} \end{gather*}(E.37)rv=(CsinθcosθCrcos2θ),(E.38)θv=(Crsin2θCcosθsinθ).
(d) One option is to compute
(E.39) ( μ v ) ν = Λ μ α Λ β ν ( α v ) β (E.39) μ v ν = Λ μ α Λ β ν α v β {:(E.39)(grad_(mu^('))v)^(nu^('))=Lambda_(mu^('))^(alpha)Lambda_(beta)^(nu^('))(grad_(alpha)v)^(beta):}\begin{equation*} \left(\boldsymbol{\nabla}_{\mu^{\prime}} \boldsymbol{v}\right)^{\nu^{\prime}}=\Lambda_{\mu^{\prime}}^{\alpha} \Lambda_{\beta}^{\nu^{\prime}}\left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{v}\right)^{\beta} \tag{E.39} \end{equation*}(E.39)(μv)ν=ΛμαΛβν(αv)β
for μ μ mu^(')\mu^{\prime}μ and ν ν nu^(')\nu^{\prime}ν each taking x x xxx and y y yyy. The only non-zero component is
( x v ) y = Λ α x Λ y β ( α v ) β = Λ r x Λ y r ( r v ) r + Λ θ x Λ y r ( θ v ) r + Λ r x Λ y θ ( r v ) θ + Λ θ x Λ y θ ( θ v ) θ = C x v y = Λ α x Λ y β α v β = Λ r x Λ y r r v r + Λ θ x Λ y r θ v r + Λ r x Λ y θ r v θ + Λ θ x Λ y θ θ v θ = C {:[(grad_(x)v)^(y)=Lambda^(alpha)_(x)Lambda^(y)_(beta)(grad_(alpha)v)^(beta)],[=Lambda^(r)_(x)Lambda^(y)_(r)(grad_(r)v)^(r)+Lambda^(theta)_(x)Lambda^(y)_(r)(grad_(theta)v)^(r)],[+Lambda^(r)_(x)Lambda^(y)_(theta)(grad_(r)v)^(theta)+Lambda^(theta)_(x)Lambda^(y)_(theta)(grad_(theta)v)^(theta)],[=C]:}\begin{aligned} \left(\boldsymbol{\nabla}_{x} \boldsymbol{v}\right)^{y} & =\Lambda^{\alpha}{ }_{x} \Lambda^{y}{ }_{\beta}\left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{v}\right)^{\beta} \\ & =\Lambda^{r}{ }_{x} \Lambda^{y}{ }_{r}\left(\boldsymbol{\nabla}_{r} \boldsymbol{v}\right)^{r}+\Lambda^{\theta}{ }_{x} \Lambda^{y}{ }_{r}\left(\boldsymbol{\nabla}_{\theta} \boldsymbol{v}\right)^{r} \\ & +\Lambda^{r}{ }_{x} \Lambda^{y}{ }_{\theta}\left(\boldsymbol{\nabla}_{r} \boldsymbol{v}\right)^{\theta}+\Lambda^{\theta}{ }_{x} \Lambda^{y}{ }_{\theta}\left(\boldsymbol{\nabla}_{\theta} \boldsymbol{v}\right)^{\theta} \\ & =C \end{aligned}(xv)y=ΛαxΛyβ(αv)β=ΛrxΛyr(rv)r+ΛθxΛyr(θv)r+ΛrxΛyθ(rv)θ+ΛθxΛyθ(θv)θ=C
(E.40)
However, it's simplest to work in the other direction and compute ( ν v ) μ = Λ x ν , Λ μ y ( x v ) y ν v μ = Λ x ν , Λ μ y x v y (grad_(nu^('))v)^(mu^('))=Lambda^(x)_(nu^(')),Lambda^(mu^('))_(y)(grad_(x)v)^(y)\left(\boldsymbol{\nabla}_{\nu^{\prime}} \boldsymbol{v}\right)^{\mu^{\prime}}=\Lambda^{x}{ }_{\nu^{\prime}}, \Lambda^{\mu^{\prime}}{ }_{y}\left(\boldsymbol{\nabla}_{x} \boldsymbol{v}\right)^{y}(νv)μ=Λxν,Λμy(xv)y for μ , ν = r , θ μ , ν = r , θ mu^('),nu^(')=r,theta\mu^{\prime}, \nu^{\prime}=r, \thetaμ,ν=r,θ and check against the results in (c).
(8.1) (a) The E-L procedure gives
(E.41) L r = r L ( d θ d λ ) 2 , L θ = 0 . L ( d r d λ ) = 1 L d r d λ , L ( d θ d λ ) = r 2 L d θ d λ . (E.41) L r = r L d θ d λ 2 , L θ = 0 . L d r d λ = 1 L d r d λ , L d θ d λ = r 2 L d θ d λ . {:(E.41){:[(del L)/(del r)=(r)/(L)(((d)theta)/((d)lambda))^(2)","quad(del L)/(del theta)=0.],[(del L)/(del(((d)r)/((d)lambda)))=(1)/(L)((d)r)/((d)lambda)","quad(del L)/(del(((d)theta)/((d)lambda)))=(r^(2))/(L)((d)theta)/((d)lambda).]:}:}\begin{array}{r} \frac{\partial L}{\partial r}=\frac{r}{L}\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right)^{2}, \quad \frac{\partial L}{\partial \theta}=0 . \\ \frac{\partial L}{\partial\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)}=\frac{1}{L} \frac{\mathrm{~d} r}{\mathrm{~d} \lambda}, \quad \frac{\partial L}{\partial\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right)}=\frac{r^{2}}{L} \frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda} . \tag{E.41} \end{array}(E.41)Lr=rL( dθ dλ)2,Lθ=0.L( dr dλ)=1L dr dλ,L( dθ dλ)=r2L dθ dλ.
At this stage we choose length parametrization
(E.42) L = [ ( d r d λ ) 2 + r 2 ( d θ d λ ) 2 ] 1 2 = 1 (E.42) L = d r d λ 2 + r 2 d θ d λ 2 1 2 = 1 {:(E.42)L=[((dr)/((d)lambda))^(2)+r^(2)(((d)theta)/((d)lambda))^(2)]^((1)/(2))=1:}\begin{equation*} L=\left[\left(\frac{\mathrm{d} r}{\mathrm{~d} \lambda}\right)^{2}+r^{2}\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right)^{2}\right]^{\frac{1}{2}}=1 \tag{E.42} \end{equation*}(E.42)L=[(dr dλ)2+r2( dθ dλ)2]12=1
We obtain two equations,
and
(E.43) d d λ L ( d r d λ ) = d 2 r d λ 2 (E.44) d d λ L ( d θ d λ ) = d d λ ( r 2 d θ d λ ) (E.43) d d λ L d r d λ = d 2 r d λ 2 (E.44) d d λ L d θ d λ = d d λ r 2 d θ d λ {:[(E.43)(d)/((d)lambda)(del L)/(del(((d)r)/((d)lambda)))=(d^(2)r)/((d)lambda^(2))],[(E.44)((d))/((d)lambda)(del L)/(del(((d)theta)/((d)lambda)))=(d)/((d)lambda)*(r^(2)((d)theta)/((d)lambda))]:}\begin{gather*} \frac{\mathrm{d}}{\mathrm{~d} \lambda} \frac{\partial L}{\partial\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)}=\frac{\mathrm{d}^{2} r}{\mathrm{~d} \lambda^{2}} \tag{E.43}\\ \frac{\mathrm{~d}}{\mathrm{~d} \lambda} \frac{\partial L}{\partial\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right)}=\frac{\mathrm{d}}{\mathrm{~d} \lambda} \cdot\left(r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right) \tag{E.44} \end{gather*}(E.43)d dλL( dr dλ)=d2r dλ2(E.44) d dλL( dθ dλ)=d dλ(r2 dθ dλ)
The two equations of motion follow.
(b) We can integrate the second of these equations to find
(E.45) r 2 d θ d λ = a (E.45) r 2 d θ d λ = a {:(E.45)r^(2)((d)theta)/((d)lambda)=a:}\begin{equation*} r^{2} \frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}=a \tag{E.45} \end{equation*}(E.45)r2 dθ dλ=a
where a a aaa is a constant. Inserting this into the length parametrization eqn (E.42) we have
(E.46) ( d r d λ ) 2 + a 2 r 2 = 1 (E.46) d r d λ 2 + a 2 r 2 = 1 {:(E.46)((dr)/((d)lambda))^(2)+(a^(2))/(r^(2))=1:}\begin{equation*} \left(\frac{\mathrm{d} r}{\mathrm{~d} \lambda}\right)^{2}+\frac{a^{2}}{r^{2}}=1 \tag{E.46} \end{equation*}(E.46)(dr dλ)2+a2r2=1
whose solution is
(E.47) r 2 = λ 2 + a 2 (E.47) r 2 = λ 2 + a 2 {:(E.47)r^(2)=lambda^(2)+a^(2):}\begin{equation*} r^{2}=\lambda^{2}+a^{2} \tag{E.47} \end{equation*}(E.47)r2=λ2+a2
Returning to the first equation of motion we find
(E.48) d θ d λ = a λ 2 + a 2 (E.48) d θ d λ = a λ 2 + a 2 {:(E.48)(dtheta)/((d)lambda)=(a)/(lambda^(2)+a^(2)):}\begin{equation*} \frac{\mathrm{d} \theta}{\mathrm{~d} \lambda}=\frac{a}{\lambda^{2}+a^{2}} \tag{E.48} \end{equation*}(E.48)dθ dλ=aλ2+a2
whose solution is
(E.49) a tan ( θ θ 0 ) = λ (E.49) a tan θ θ 0 = λ {:(E.49)a tan(theta-theta_(0))=lambda:}\begin{equation*} a \tan \left(\theta-\theta_{0}\right)=\lambda \tag{E.49} \end{equation*}(E.49)atan(θθ0)=λ
(8.2) (b) Using x = r cos θ x = r cos θ x=r cos thetax=r \cos \thetax=rcosθ and y = r sin θ y = r sin θ y=r sin thetay=r \sin \thetay=rsinθ we have
(E.50) r = γ α cos θ + β sin θ . (E.50) r = γ α cos θ + β sin θ . {:(E.50)r=(gamma)/(alpha cos theta+beta sin theta).:}\begin{equation*} r=\frac{\gamma}{\alpha \cos \theta+\beta \sin \theta} . \tag{E.50} \end{equation*}(E.50)r=γαcosθ+βsinθ.
If we make the substitutions
a = γ ( α 2 + β 2 ) 1 2 , cos θ 0 = α ( α 2 + β 2 ) 1 2 , sin θ 0 = β ( α 2 + β 2 ) 1 2 , a = γ α 2 + β 2 1 2 , cos θ 0 = α α 2 + β 2 1 2 , sin θ 0 = β α 2 + β 2 1 2 , {:[a=(gamma)/((alpha^(2)+beta^(2))^((1)/(2)))","quad cos theta_(0)=(alpha)/((alpha^(2)+beta^(2))^((1)/(2)))","],[sin theta_(0)=(beta)/((alpha^(2)+beta^(2))^((1)/(2)))","]:}\begin{aligned} & a=\frac{\gamma}{\left(\alpha^{2}+\beta^{2}\right)^{\frac{1}{2}}}, \quad \cos \theta_{0}=\frac{\alpha}{\left(\alpha^{2}+\beta^{2}\right)^{\frac{1}{2}}}, \\ & \sin \theta_{0}=\frac{\beta}{\left(\alpha^{2}+\beta^{2}\right)^{\frac{1}{2}}}, \end{aligned}a=γ(α2+β2)12,cosθ0=α(α2+β2)12,sinθ0=β(α2+β2)12,
then eqn 8.48 becomes r = a / cos ( θ θ 0 ) r = a / cos θ θ 0 r=a//cos(theta-theta_(0))r=a / \cos \left(\theta-\theta_{0}\right)r=a/cos(θθ0), which shows that this is, indeed, a description of a straight line.
(8.3) First evaluate velocity
(E.51) d ξ μ d τ = ξ μ x ν d x ν d τ (E.51) d ξ μ d τ = ξ μ x ν d x ν d τ {:(E.51)(dxi^(mu))/(dtau)=(delxi^(mu))/(delx^(nu))*(dx^(nu))/(dtau):}\begin{equation*} \frac{\mathrm{d} \xi^{\mu}}{\mathrm{d} \tau}=\frac{\partial \xi^{\mu}}{\partial x^{\nu}} \cdot \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} \tag{E.51} \end{equation*}(E.51)dξμdτ=ξμxνdxνdτ
This allows us to evaluate the acceleration
d 2 ξ μ d τ 2 = d x σ d τ x σ ξ μ x ν d x ν d τ d 2 ξ μ d τ 2 = d x σ d τ x σ ξ μ x ν d x ν d τ (d^(2)xi^(mu))/(dtau^(2))=(dx^(sigma))/(dtau)*(del)/(delx^(sigma))(delxi^(mu))/(delx^(nu))*(dx^(nu))/(dtau)\frac{\mathrm{d}^{2} \xi^{\mu}}{\mathrm{d} \tau^{2}}=\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial}{\partial x^{\sigma}} \frac{\partial \xi^{\mu}}{\partial x^{\nu}} \cdot \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}d2ξμdτ2=dxσdτxσξμxνdxνdτ
= d x σ d τ 2 ξ μ x ν x σ d x ν d τ + d x σ d τ ξ μ x ν x σ d x ν d τ = d x σ d τ 2 ξ μ x ν x σ d x ν d τ + d x σ d τ ξ μ x ν x σ d x ν d τ =(dx^(sigma))/(dtau)*(del^(2)xi^(mu))/(delx^(nu)delx^(sigma))(dx^(nu))/(dtau)+(dx^(sigma))/(dtau)*(delxi^(mu))/(delx^(nu))(del)/(delx^(sigma))(dx^(nu))/(dtau)=\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial^{2} \xi^{\mu}}{\partial x^{\nu} \partial x^{\sigma}} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}+\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial \xi^{\mu}}{\partial x^{\nu}} \frac{\partial}{\partial x^{\sigma}} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}=dxσdτ2ξμxνxσdxνdτ+dxσdτξμxνxσdxνdτ
= d x σ d τ 2 ξ μ x λ x σ d x λ d τ + ξ μ x ν d 2 x ν d τ 2 = 0 , = d x σ d τ 2 ξ μ x λ x σ d x λ d τ + ξ μ x ν d 2 x ν d τ 2 = 0 , =(dx^(sigma))/(dtau)*(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma))*(dx^(lambda))/(dtau)+(delxi^(mu))/(delx^(nu))(d^(2)x^(nu))/(dtau^(2))=0,quad=\frac{\mathrm{d} x^{\sigma}}{\mathrm{d} \tau} \cdot \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}} \cdot \frac{\mathrm{d} x^{\lambda}}{\mathrm{d} \tau}+\frac{\partial \xi^{\mu}}{\partial x^{\nu}} \frac{\mathrm{d}^{2} x^{\nu}}{\mathrm{d} \tau^{2}}=0, \quad=dxσdτ2ξμxλxσdxλdτ+ξμxνd2xνdτ2=0, (E.52)
where, in the first term of the last line, we sum over λ λ lambda\lambdaλ instead of ν ν nu\nuν to avoid confusion in the next stage. Tidying, we have
(E.53) d 2 x ν d τ 2 + x ν ξ μ 2 ξ μ x λ x σ d x λ d τ d x σ d τ = 0 (E.53) d 2 x ν d τ 2 + x ν ξ μ 2 ξ μ x λ x σ d x λ d τ d x σ d τ = 0 {:(E.53)(d^(2)x^(nu))/(dtau^(2))+(delx^(nu))/(delxi^(mu))(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma))(dx^(lambda))/(dtau)*((d)x^(sigma))/(dtau)=0:}\begin{equation*} \frac{\mathrm{d}^{2} x^{\nu}}{\mathrm{d} \tau^{2}}+\frac{\partial x^{\nu}}{\partial \xi^{\mu}} \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}} \frac{\mathrm{d} x^{\lambda}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} x^{\sigma}}{\mathrm{d} \tau}=0 \tag{E.53} \end{equation*}(E.53)d2xνdτ2+xνξμ2ξμxλxσdxλdτ dxσdτ=0
This is
(E.54) d 2 x ν d τ 2 + Γ λ σ ν d x λ d τ d x σ d τ = 0 (E.54) d 2 x ν d τ 2 + Γ λ σ ν d x λ d τ d x σ d τ = 0 {:(E.54)(d^(2)x^(nu))/(dtau^(2))+Gamma_(lambda sigma)^(nu)(dx^(lambda))/(dtau)((d)x^(sigma))/(dtau)=0:}\begin{equation*} \frac{\mathrm{d}^{2} x^{\nu}}{\mathrm{d} \tau^{2}}+\Gamma_{\lambda \sigma}^{\nu} \frac{\mathrm{d} x^{\lambda}}{\mathrm{d} \tau} \frac{\mathrm{~d} x^{\sigma}}{\mathrm{d} \tau}=0 \tag{E.54} \end{equation*}(E.54)d2xνdτ2+Γλσνdxλdτ dxσdτ=0
with
(E.55) Γ λ σ ν = x ν ξ μ 2 ξ μ x λ x σ (E.55) Γ λ σ ν = x ν ξ μ 2 ξ μ x λ x σ {:(E.55)Gamma_(lambda sigma)^(nu)=(delx^(nu))/(delxi^(mu))(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma)):}\begin{equation*} \Gamma_{\lambda \sigma}^{\nu}=\frac{\partial x^{\nu}}{\partial \xi^{\mu}} \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}} \tag{E.55} \end{equation*}(E.55)Γλσν=xνξμ2ξμxλxσ
(8.4) (a) Following the same steps yields
(E.56) ( u u ¯ ) μ = u α ( u μ x α u λ Γ α μ λ ) = 0 , (E.56) u u ¯ μ = u α u μ x α u λ Γ α μ λ = 0 , {:(E.56)(grad_(u)( bar(u)))_(mu)=u^(alpha)((delu_(mu))/(delx^(alpha))-u_(lambda)Gamma_(alpha mu)^(lambda))=0",":}\begin{equation*} \left(\nabla_{u} \bar{u}\right)_{\mu}=u^{\alpha}\left(\frac{\partial u_{\mu}}{\partial x^{\alpha}}-u_{\lambda} \Gamma_{\alpha \mu}^{\lambda}\right)=0, \tag{E.56} \end{equation*}(E.56)(uu¯)μ=uα(uμxαuλΓαμλ)=0,
and so
(E.57) x ¨ μ Γ μ α λ x ˙ α x ˙ λ = 0 (E.57) x ¨ μ Γ μ α λ x ˙ α x ˙ λ = 0 {:(E.57)x^(¨)_(mu)-Gamma_(mu alpha)^(lambda)x^(˙)^(alpha)x^(˙)_(lambda)=0:}\begin{equation*} \ddot{x}_{\mu}-\Gamma_{\mu \alpha}^{\lambda} \dot{x}^{\alpha} \dot{x}_{\lambda}=0 \tag{E.57} \end{equation*}(E.57)x¨μΓμαλx˙αx˙λ=0
(b) Start with u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1, which implies d / d τ ( u u ) = 0 d / d τ ( u u ) = 0 d//dtau(u*u)=0\mathrm{d} / \mathrm{d} \tau(\boldsymbol{u} \cdot \boldsymbol{u})=0d/dτ(uu)=0, from which we say
(E.58) u ˙ μ u μ + u μ u ˙ μ = 0 (E.58) u ˙ μ u μ + u μ u ˙ μ = 0 {:(E.58)u^(˙)^(mu)u_(mu)+u^(mu)u^(˙)_(mu)=0:}\begin{equation*} \dot{u}^{\mu} u_{\mu}+u^{\mu} \dot{u}_{\mu}=0 \tag{E.58} \end{equation*}(E.58)u˙μuμ+uμu˙μ=0
Now consider the equations of motion dotted with the velocity, in the forms
u μ u ˙ μ + u μ Γ α β μ u α u β = f μ u μ / m (E.59) u μ u ˙ μ u μ Γ α μ λ u α u λ = f μ u μ / m u μ u ˙ μ + u μ Γ α β μ u α u β = f μ u μ / m (E.59) u μ u ˙ μ u μ Γ α μ λ u α u λ = f μ u μ / m {:[u_(mu)u^(˙)^(mu)+u_(mu)Gamma_(alpha beta)^(mu)u^(alpha)u^(beta)=f^(mu)u_(mu)//m],[(E.59)u^(mu)u^(˙)_(mu)-u^(mu)Gamma_(alpha mu)^(lambda)u^(alpha)u_(lambda)=f_(mu)u^(mu)//m]:}\begin{align*} u_{\mu} \dot{u}^{\mu}+u_{\mu} \Gamma_{\alpha \beta}^{\mu} u^{\alpha} u^{\beta} & =f^{\mu} u_{\mu} / m \\ u^{\mu} \dot{u}_{\mu}-u^{\mu} \Gamma_{\alpha \mu}^{\lambda} u^{\alpha} u_{\lambda} & =f_{\mu} u^{\mu} / m \tag{E.59} \end{align*}uμu˙μ+uμΓαβμuαuβ=fμuμ/m(E.59)uμu˙μuμΓαμλuαuλ=fμuμ/m
Add these together to see that the right-hand side must vanish.
(9.4) (a) The Lagrangian, parametrized by the proper time,
(E.60) L = [ x 2 ( d t d τ ) 2 ( d x d τ ) 2 ] 1 2 (E.60) L = x 2 d t d τ 2 d x d τ 2 1 2 {:(E.60)L=[x^(2)(((d)t)/((d)tau))^(2)-((dx)/((d)tau))^(2)]^((1)/(2)):}\begin{equation*} L=\left[x^{2}\left(\frac{\mathrm{~d} t}{\mathrm{~d} \tau}\right)^{2}-\left(\frac{\mathrm{d} x}{\mathrm{~d} \tau}\right)^{2}\right]^{\frac{1}{2}} \tag{E.60} \end{equation*}(E.60)L=[x2( dt dτ)2(dx dτ)2]12
Feeding this into the Euler-Lagrange equations we find
L t = x 2 L t ˙ , L x ˙ = 1 L x ˙ , L x = 1 L x t ˙ 2 L t = x 2 L t ˙ , L x ˙ = 1 L x ˙ , L x = 1 L x t ˙ 2 (del L)/(del t)=(x^(2))/(L)*t^(˙),quad(del L)/(del(x^(˙)))=-(1)/(L)x^(˙),quad(del L)/(del x)=(1)/(L)xt^(˙)^(2)\frac{\partial L}{\partial t}=\frac{x^{2}}{L} \cdot \dot{t}, \quad \frac{\partial L}{\partial \dot{x}}=-\frac{1}{L} \dot{x}, \quad \frac{\partial L}{\partial x}=\frac{1}{L} x \dot{t}^{2}Lt=x2Lt˙,Lx˙=1Lx˙,Lx=1Lxt˙2
We now choose length parametrization, which sets L = L = L=L=L= 1 , and so
(E.62) L 2 = x 2 ( d t d τ ) 2 ( d x d τ ) 2 = 1 (E.62) L 2 = x 2 d t d τ 2 d x d τ 2 = 1 {:(E.62)L^(2)=x^(2)(((d)t)/((d)tau))^(2)-((dx)/((d)tau))^(2)=1:}\begin{equation*} L^{2}=x^{2}\left(\frac{\mathrm{~d} t}{\mathrm{~d} \tau}\right)^{2}-\left(\frac{\mathrm{d} x}{\mathrm{~d} \tau}\right)^{2}=1 \tag{E.62} \end{equation*}(E.62)L2=x2( dt dτ)2(dx dτ)2=1
The resulting equations of motion are
(E.63) 2 x ˙ t ˙ + x t ¨ = 0 , x ¨ + x t ˙ 2 = 0 (E.63) 2 x ˙ t ˙ + x t ¨ = 0 , x ¨ + x t ˙ 2 = 0 {:(E.63)2x^(˙)t^(˙)+xt^(¨)=0","quadx^(¨)+xt^(˙)^(2)=0:}\begin{equation*} 2 \dot{x} \dot{t}+x \ddot{t}=0, \quad \ddot{x}+x \dot{t}^{2}=0 \tag{E.63} \end{equation*}(E.63)2x˙t˙+xt¨=0,x¨+xt˙2=0
The connection coefficients are Γ t x t = 1 / x Γ t x t = 1 / x Gamma^(t)_(xt)=1//x\Gamma^{t}{ }_{x t}=1 / xΓtxt=1/x and Γ x t t = x Γ x t t = x Gamma^(x)_(tt)=x\Gamma^{x}{ }_{t t}=xΓxtt=x. (b) Using the chain rule we write
(E.64) d x d τ = d x d t d t d τ (E.64) d x d τ = d x d t d t d τ {:(E.64)(dx)/((d)tau)=(dx)/((d)t)((d)t)/((d)tau):}\begin{equation*} \frac{\mathrm{d} x}{\mathrm{~d} \tau}=\frac{\mathrm{d} x}{\mathrm{~d} t} \frac{\mathrm{~d} t}{\mathrm{~d} \tau} \tag{E.64} \end{equation*}(E.64)dx dτ=dx dt dt dτ
The condition L = 1 L = 1 L=1L=1L=1 becomes
(E.65) ( x 2 v 2 ) t ˙ 2 = 1 t ˙ = ( x 2 v 2 ) 1 2 (E.65) x 2 v 2 t ˙ 2 = 1 t ˙ = x 2 v 2 1 2 {:[(E.65)(x^(2)-v^(2))t^(˙)^(2)=1],[t^(˙)=(x^(2)-v^(2))^(-(1)/(2))]:}\begin{align*} & \left(x^{2}-v^{2}\right) \dot{t}^{2}=1 \tag{E.65}\\ & \dot{t}=\left(x^{2}-v^{2}\right)^{-\frac{1}{2}} \end{align*}(E.65)(x2v2)t˙2=1t˙=(x2v2)12
(E.66)
Substituting this into the second of the equations of motion, we find
(E.67) x ¨ = x x 2 v 2 . (E.67) x ¨ = x x 2 v 2 . {:(E.67)x^(¨)=-(x)/(x^(2)-v^(2)).:}\begin{equation*} \ddot{x}=-\frac{x}{x^{2}-v^{2}} . \tag{E.67} \end{equation*}(E.67)x¨=xx2v2.
(9.5) (a) We have the inverse matrix
(E.68) g μ ν = ( 1 Ω y Ω x 0 Ω y 1 Ω 2 y 2 Ω 2 x y 0 Ω x Ω 2 x y 1 Ω 2 x 2 0 0 0 0 1 ) (E.68) g μ ν = 1 Ω y Ω x 0 Ω y 1 Ω 2 y 2 Ω 2 x y 0 Ω x Ω 2 x y 1 Ω 2 x 2 0 0 0 0 1 {:(E.68)g^(mu nu)=([-1,-Omega y,Omega x,0],[-Omega y,1-Omega^(2)y^(2),Omega^(2)xy,0],[Omega x,Omega^(2)xy,1-Omega^(2)x^(2),0],[0,0,0,1]):}g^{\mu \nu}=\left(\begin{array}{cccc} -1 & -\Omega y & \Omega x & 0 \tag{E.68}\\ -\Omega y & 1-\Omega^{2} y^{2} & \Omega^{2} x y & 0 \\ \Omega x & \Omega^{2} x y & 1-\Omega^{2} x^{2} & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)(E.68)gμν=(1ΩyΩx0Ωy1Ω2y2Ω2xy0ΩxΩ2xy1Ω2x200001)
(b) The non-zero connection coefficients are
(E.69) Γ x t t = Ω 2 x , Γ y t t = Ω 2 y Γ x y t Ω , Γ y x t = Ω (E.69) Γ x t t = Ω 2 x , Γ y t t = Ω 2 y Γ x y t Ω , Γ y x t = Ω {:(E.69){:[Gamma^(x)_(tt)=-Omega^(2)x",",Gamma^(y)_(tt)=-Omega^(2)y],[Gamma^(x)_(yt)-Omega",",Gamma^(y)_(xt)=Omega]:}:}\begin{array}{cc} \Gamma^{x}{ }_{t t}=-\Omega^{2} x, & \Gamma^{y}{ }_{t t}=-\Omega^{2} y \tag{E.69}\\ \Gamma^{x}{ }_{y t}-\Omega, & \Gamma^{y}{ }_{x t}=\Omega \end{array}(E.69)Γxtt=Ω2x,Γytt=Ω2yΓxytΩ,Γyxt=Ω
(9.6) (a) The rotating-frame line element is
d s 2 = d t 2 ( 1 Ω 2 r 2 ) + d r 2 + r 2 d θ 2 (E.70) + d z 2 + 2 Ω r 2 d θ d t d s 2 = d t 2 1 Ω 2 r 2 + d r 2 + r 2 d θ 2 (E.70) + d z 2 + 2 Ω r 2 d θ d t {:[ds^(2)=-dt^(2)(1-Omega^(2)r^(2))+dr^(2)+r^(2)dtheta^(2)],[(E.70)+dz^(2)+2Omegar^(2)dthetadt]:}\begin{align*} \mathrm{d} s^{2} & =-\mathrm{d} t^{2}\left(1-\Omega^{2} r^{2}\right)+\mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2} \\ & +\mathrm{d} z^{2}+2 \Omega r^{2} \mathrm{~d} \theta \mathrm{~d} t \tag{E.70} \end{align*}ds2=dt2(1Ω2r2)+dr2+r2 dθ2(E.70)+dz2+2Ωr2 dθ dt
(b) The non-zero connection coefficients are
(E.71) Γ r θ θ = r , Γ r θ t = Ω r , Γ r θ θ = 1 r , Γ r t θ = Ω r . (E.71) Γ r θ θ = r , Γ r θ t = Ω r , Γ r θ θ = 1 r , Γ r t θ = Ω r . {:(E.71){:[Gamma^(r)_(theta theta)=-r","quadGamma^(r)_(theta t)=-Omega r","],[Gamma_(r theta)^(theta)=(1)/(r)",",Gamma_(rt)^(theta)=(Omega )/(r).]:}:}\begin{array}{cc} \Gamma^{r}{ }_{\theta \theta}=-r, \quad \Gamma^{r}{ }_{\theta t}=-\Omega r, \tag{E.71}\\ \Gamma_{r \theta}^{\theta}=\frac{1}{r}, & \Gamma_{r t}^{\theta}=\frac{\Omega}{r} . \end{array}(E.71)Γrθθ=r,Γrθt=Ωr,Γrθθ=1r,Γrtθ=Ωr.
(9.7) We find
(E.72) L t = e 2 Φ 1 L d t d λ , L r = e 2 Λ 1 L d r d λ , L θ = r 2 1 L d θ d λ , L ϕ = r 2 sin 2 θ 1 L d ϕ d λ . (E.72) L t = e 2 Φ 1 L d t d λ , L r = e 2 Λ 1 L d r d λ , L θ = r 2 1 L d θ d λ , L ϕ = r 2 sin 2 θ 1 L d ϕ d λ . {:(E.72){:[(del L)/(del t)=e^(2Phi)(1)/(L)((d)t)/((d)lambda)",",(del L)/(del r)=-e^(2Lambda)(1)/(L)((d)r)/((d)lambda)","],[(del L)/(del theta)=-r^(2)(1)/(L)((d)theta)/((d)lambda)",",(del L)/(del phi)=-r^(2)sin^(2)theta(1)/(L)((d)phi)/((d)lambda).]:}:}\begin{array}{lc} \frac{\partial L}{\partial t}=\mathrm{e}^{2 \Phi} \frac{1}{L} \frac{\mathrm{~d} t}{\mathrm{~d} \mathrm{\lambda}}, & \frac{\partial L}{\partial r}=-\mathrm{e}^{2 \Lambda} \frac{1}{L} \frac{\mathrm{~d} r}{\mathrm{~d} \lambda}, \tag{E.72}\\ \frac{\partial L}{\partial \theta}=-r^{2} \frac{1}{L} \frac{\mathrm{~d} \theta}{\mathrm{~d} \mathrm{\lambda}}, & \frac{\partial L}{\partial \phi}=-r^{2} \sin ^{2} \theta \frac{1}{L} \frac{\mathrm{~d} \phi}{\mathrm{~d} \lambda} . \end{array}(E.72)Lt=e2Φ1L dt dλ,Lr=e2Λ1L dr dλ,Lθ=r21L dθ dλ,Lϕ=r2sin2θ1L dϕ dλ.
This leads to E-L equation 1
(E.73) d 2 t d λ 2 + 2 Φ r d r d λ d t d λ = 0 (E.73) d 2 t d λ 2 + 2 Φ r d r d λ d t d λ = 0 {:(E.73)(d^(2)t)/((d)lambda^(2))+2(del Phi)/(del r)((d)r)/((d)lambda)((d)t)/((d)lambda)=0:}\begin{equation*} \frac{\mathrm{d}^{2} t}{\mathrm{~d} \lambda^{2}}+2 \frac{\partial \Phi}{\partial r} \frac{\mathrm{~d} r}{\mathrm{~d} \lambda} \frac{\mathrm{~d} t}{\mathrm{~d} \lambda}=0 \tag{E.73} \end{equation*}(E.73)d2t dλ2+2Φr dr dλ dt dλ=0
giving Γ t r t = Φ r Γ t r t = Φ r Gamma^(t)_(rt)=(del Phi)/(del r)\Gamma^{t}{ }_{r t}=\frac{\partial \Phi}{\partial r}Γtrt=Φr. E-L equation 2 is
d 2 r d λ 2 + Φ r e 2 Φ e 2 Λ ( d t d λ ) 2 + Λ r ( d r d λ ) 2 d 2 r d λ 2 + Φ r e 2 Φ e 2 Λ d t d λ 2 + Λ r d r d λ 2 (d^(2)r)/((d)lambda^(2))+(del Phi)/(del r)e^(2Phi)e^(-2Lambda)(((d)t)/((d)lambda))^(2)+(del Lambda)/(del r)(((d)r)/((d)lambda))^(2)\frac{\mathrm{d}^{2} r}{\mathrm{~d} \lambda^{2}}+\frac{\partial \Phi}{\partial r} \mathrm{e}^{2 \Phi} \mathrm{e}^{-2 \Lambda}\left(\frac{\mathrm{~d} t}{\mathrm{~d} \lambda}\right)^{2}+\frac{\partial \Lambda}{\partial r}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \lambda}\right)^{2}d2r dλ2+Φre2Φe2Λ( dt dλ)2+Λr( dr dλ)2
r e 2 Λ ( d θ d λ ) 2 r e 2 Λ sin 2 θ ( d ϕ d λ ) 2 = 0 , (E.74) r e 2 Λ d θ d λ 2 r e 2 Λ sin 2 θ d ϕ d λ 2 = 0 ,  (E.74)  -re^(-2Lambda)(((d)theta)/((d)lambda))^(2)-re^(-2Lambda)sin^(2)theta(((d)phi)/((d)lambda))^(2)=0,quad" (E.74) "-r \mathrm{e}^{-2 \Lambda}\left(\frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}\right)^{2}-r \mathrm{e}^{-2 \Lambda} \sin ^{2} \theta\left(\frac{\mathrm{~d} \phi}{\mathrm{~d} \lambda}\right)^{2}=0, \quad \text { (E.74) }re2Λ( dθ dλ)2re2Λsin2θ( dϕ dλ)2=0, (E.74) 
giving
(E.75) Γ t t r = Φ e 2 Φ e 2 Λ , Γ r r r = Λ , Γ r θ θ = r e 2 Λ , Γ ϕ ϕ r = r e 2 Λ sin 2 θ (E.75) Γ t t r = Φ e 2 Φ e 2 Λ , Γ r r r = Λ , Γ r θ θ = r e 2 Λ , Γ ϕ ϕ r = r e 2 Λ sin 2 θ {:[(E.75)Gamma_(tt)^(r)=Phi^(')e^(2Phi)e^(-2Lambda)","quadGamma_(rr)^(r)=Lambda^(')","],[Gamma^(r)_(theta theta)=-re^(-2Lambda)","quadGamma_(phi phi)^(r)=-re^(-2Lambda)sin^(2)theta]:}\begin{gather*} \Gamma_{t t}^{r}=\Phi^{\prime} \mathrm{e}^{2 \Phi} \mathrm{e}^{-2 \Lambda}, \quad \Gamma_{r r}^{r}=\Lambda^{\prime}, \tag{E.75}\\ \Gamma^{r}{ }_{\theta \theta}=-r \mathrm{e}^{-2 \Lambda}, \quad \Gamma_{\phi \phi}^{r}=-r \mathrm{e}^{-2 \Lambda} \sin ^{2} \theta \end{gather*}(E.75)Γttr=Φe2Φe2Λ,Γrrr=Λ,Γrθθ=re2Λ,Γϕϕr=re2Λsin2θ
E-L equation 3 is
d 2 θ d λ 2 + 2 r d r d λ d θ d λ sin θ cos θ ( d ϕ d λ ) 2 = 0 giving (E.77) Γ θ r θ = 1 r , Γ θ ϕ ϕ = sin θ cos θ d 2 θ d λ 2 + 2 r d r d λ d θ d λ sin θ cos θ d ϕ d λ 2 = 0  giving  (E.77) Γ θ r θ = 1 r , Γ θ ϕ ϕ = sin θ cos θ {:[(d^(2)theta)/((d)lambda^(2))+(2)/(r)*((d)r)/((d)lambda)((d)theta)/((d)lambda)-sin theta cos theta((dphi)/((d)lambda))^(2)=0],[" giving "],[(E.77)Gamma^(theta)_(r theta)=(1)/(r)","quadGamma^(theta)_(phi phi)=-sin theta cos theta]:}\begin{gather*} \frac{\mathrm{d}^{2} \theta}{\mathrm{~d} \lambda^{2}}+\frac{2}{r} \cdot \frac{\mathrm{~d} r}{\mathrm{~d} \lambda} \frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda}-\sin \theta \cos \theta\left(\frac{\mathrm{d} \phi}{\mathrm{~d} \lambda}\right)^{2}=0 \\ \text { giving } \\ \Gamma^{\theta}{ }_{r \theta}=\frac{1}{r}, \quad \Gamma^{\theta}{ }_{\phi \phi}=-\sin \theta \cos \theta \tag{E.77} \end{gather*}d2θ dλ2+2r dr dλ dθ dλsinθcosθ(dϕ dλ)2=0 giving (E.77)Γθrθ=1r,Γθϕϕ=sinθcosθ
E-L equation 4 is
(E.78) d 2 ϕ d λ 2 + 2 r d r d λ d ϕ d λ + 2 cos θ sin θ 2 r d θ d λ d ϕ d λ = 0 , (E.79) Γ r ϕ ϕ = 1 r , Γ θ ϕ ϕ = cos θ sin θ . (E.78) d 2 ϕ d λ 2 + 2 r d r d λ d ϕ d λ + 2 cos θ sin θ 2 r d θ d λ d ϕ d λ = 0 , (E.79) Γ r ϕ ϕ = 1 r , Γ θ ϕ ϕ = cos θ sin θ . {:[(E.78)(d^(2)phi)/((d)lambda^(2))+(2)/(r)((d)r)/((d)lambda)*((d)phi)/((d)lambda)+2(cos theta)/(sin theta)(2)/(r)((d)theta)/((d)lambda)((d)phi)/((d)lambda)=0","],[(E.79)Gamma_(r phi)^(phi)=(1)/(r)","quadGamma_(theta phi)^(phi)=(cos theta)/(sin theta).]:}\begin{gather*} \frac{\mathrm{d}^{2} \phi}{\mathrm{~d} \lambda^{2}}+\frac{2}{r} \frac{\mathrm{~d} r}{\mathrm{~d} \lambda} \cdot \frac{\mathrm{~d} \phi}{\mathrm{~d} \lambda}+2 \frac{\cos \theta}{\sin \theta} \frac{2}{r} \frac{\mathrm{~d} \theta}{\mathrm{~d} \lambda} \frac{\mathrm{~d} \phi}{\mathrm{~d} \lambda}=0, \tag{E.78}\\ \Gamma_{r \phi}^{\phi}=\frac{1}{r}, \quad \Gamma_{\theta \phi}^{\phi}=\frac{\cos \theta}{\sin \theta} . \tag{E.79} \end{gather*}(E.78)d2ϕ dλ2+2r dr dλ dϕ dλ+2cosθsinθ2r dθ dλ dϕ dλ=0,(E.79)Γrϕϕ=1r,Γθϕϕ=cosθsinθ.
giving
(9.8) (b) Using the result from part (a) and rearranging we find
(E.80) d λ = d r r ( 1 r 2 a 2 ) 1 2 (E.80) d λ = d r r 1 r 2 a 2 1 2 {:(E.80)dlambda=(dr)/(r(1-(r^(2))/(a^(2)))^((1)/(2))):}\begin{equation*} \mathrm{d} \lambda=\frac{\mathrm{d} r}{r\left(1-\frac{r^{2}}{a^{2}}\right)^{\frac{1}{2}}} \tag{E.80} \end{equation*}(E.80)dλ=drr(1r2a2)12
The substitution r = a sin t r = a sin t r=a sin tr=a \sin tr=asint then gives the stated result.
(c) Write
(E.81) d x = r 2 a d λ (E.81) d x = r 2 a d λ {:(E.81)dx=(r^(2))/(a)*dlambda:}\begin{equation*} \mathrm{d} x=\frac{r^{2}}{a} \cdot \mathrm{~d} \lambda \tag{E.81} \end{equation*}(E.81)dx=r2a dλ
and substitute for d λ d λ dlambda\mathrm{d} \lambdadλ from part (b).
(e) The length is
Δ λ = a b d λ d t d t = a b d t sin t = ln | 1 + cos t sin t | ( E .82 ) b Δ λ = a b d λ d t d t = a b d t sin t = ln 1 + cos t sin t ( E .82 ) b Delta lambda=int_(a)^(b)((d)lambda)/((d)t)dt=int_(a)^(b)((d)t)/(sin t)=-ln |(1+cos t)/(sin t)|_((E.82))^(b)\Delta \lambda=\int_{a}^{b} \frac{\mathrm{~d} \lambda}{\mathrm{~d} t} \mathrm{~d} t=\int_{a}^{b} \frac{\mathrm{~d} t}{\sin t}=-\ln \left|\frac{1+\cos t}{\sin t}\right|_{(\mathrm{E} .82)}^{b}Δλ=ab dλ dt dt=ab dtsint=ln|1+costsint|(E.82)b
The length of the so-called maximal geodesic, that starts at t = 0 t = 0 t=0t=0t=0 and ends at t = π t = π t=pit=\pit=π, is infinite.
(9.9) Call y = ( g μ ν x ˙ μ x ˙ ν ) 1 2 y = g μ ν x ˙ μ x ˙ ν 1 2 y=(-g_(mu nu)x^(˙)^(mu)x^(˙)^(nu))^((1)/(2))y=\left(-g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}\right)^{\frac{1}{2}}y=(gμνx˙μx˙ν)12. The Euler-Lagrange equation gives
d 2 F d y 2 d y d λ y x ˙ + d F d y [ d d λ ( y x ˙ ) y x ] = 0 d 2 F d y 2 d y d λ y x ˙ + d F d y d d λ y x ˙ y x = 0 (d^(2)F)/((d)y^(2))((d)y)/((d)lambda)*(del y)/(del(x^(˙)))+(dF)/((d)y)*[((d))/((d)lambda)((del y)/(del(x^(˙))))-(del y)/(del x)]=0\frac{\mathrm{d}^{2} F}{\mathrm{~d} y^{2}} \frac{\mathrm{~d} y}{\mathrm{~d} \lambda} \cdot \frac{\partial y}{\partial \dot{x}}+\frac{\mathrm{d} F}{\mathrm{~d} y} \cdot\left[\frac{\mathrm{~d}}{\mathrm{~d} \lambda}\left(\frac{\partial y}{\partial \dot{x}}\right)-\frac{\partial y}{\partial x}\right]=0d2F dy2 dy dλyx˙+dF dy[ d dλ(yx˙)yx]=0
For length parametrization we have d y / d λ = 0 d y / d λ = 0 dy//dlambda=0\mathrm{d} y / \mathrm{d} \lambda=0dy/dλ=0, wiping out the first term. We also have that d F / d y 0 d F / d y 0 dF//dy!=0\mathrm{d} F / \mathrm{d} y \neq 0dF/dy0, as the function is monotonic. This then results in the same Euler-Lagrange equation as obtained from varying y y yyy.
(9.11) We find connection coefficients
Γ r r r = 1 2 ( 1 + r ) , Γ r ϕ ϕ = r 1 + r , Γ r ϕ ϕ r ϕ = 1 r . Γ r r r = 1 2 ( 1 + r ) , Γ r ϕ ϕ = r 1 + r , Γ r ϕ ϕ r ϕ = 1 r . Gamma^(r)_(rr)=(1)/(2(1+r)),quadGamma^(r)_(phi phi)=-(r)/(1+r),quadGamma_(r phi)^(phi)_(r phi)=(1)/(r).\Gamma^{r}{ }_{r r}=\frac{1}{2(1+r)}, \quad \Gamma^{r}{ }_{\phi \phi}=-\frac{r}{1+r}, \quad \Gamma_{r \phi}^{\phi}{ }_{r \phi}=\frac{1}{r} .Γrrr=12(1+r),Γrϕϕ=r1+r,Γrϕϕrϕ=1r.
(10.6) (a) Using η μ ^ ν = e μ ^ e ν ^ η μ ^ ν = e μ ^ e ν ^ eta_( hat(mu)nu)=e_( hat(mu))*e_( hat(nu))\eta_{\hat{\mu} \nu}=\boldsymbol{e}_{\hat{\mu}} \cdot \boldsymbol{e}_{\hat{\nu}}ημ^ν=eμ^eν^ we find
Differentiating with respect to λ λ lambda\lambdaλ, the
Differentiating with respect to λ λ lambda\lambdaλ, the velocity has components u μ = ( d r / d λ , d ϕ / d λ ) u μ = ( d r / d λ , d ϕ / d λ ) u^(mu)=(dr//dlambda,dphi//dlambda)u^{\mu}=(\mathrm{d} r / \mathrm{d} \lambda, \mathrm{d} \phi / \mathrm{d} \lambda)uμ=(dr/dλ,dϕ/dλ) given by
(E.85) u μ = ( ( 1 + r ) 1 2 , 0 ) (E.85) u μ = ( 1 + r ) 1 2 , 0 {:(E.85)u^(mu)=((1+r)^(-(1)/(2)),0):}\begin{equation*} u^{\mu}=\left((1+r)^{-\frac{1}{2}}, 0\right) \tag{E.85} \end{equation*}(E.85)uμ=((1+r)12,0)
We find
(E.86) u ; r r = u r r + Γ r r r ( u r ) 2 = 0 , (E.87) u ϕ ; ϕ = r 1 ( r + 1 ) 1 2 , (E.86) u ; r r = u r r + Γ r r r u r 2 = 0 , (E.87) u ϕ ; ϕ = r 1 ( r + 1 ) 1 2 , {:[(E.86)u_(;r)^(r)=(delu^(r))/(del r)+Gamma^(r)_(rr)(u^(r))^(2)=0","],[(E.87)u^(phi)_(;phi)=r^(-1)(r+1)^((1)/(2))","]:}\begin{align*} & u_{; r}^{r}=\frac{\partial u^{r}}{\partial r}+\Gamma^{r}{ }_{r r}\left(u^{r}\right)^{2}=0, \tag{E.86}\\ & u^{\phi}{ }_{; \phi}=r^{-1}(r+1)^{\frac{1}{2}}, \tag{E.87} \end{align*}(E.86)u;rr=urr+Γrrr(ur)2=0,(E.87)uϕ;ϕ=r1(r+1)12,
and the other components vanish. Contracting against the components of the velocity then gives a vanishing acceleration.
(10.1) (a) We obtain
A μ ^ = ( A t , a A χ , a sinh χ A θ , a sinh χ sin θ A ϕ ) A μ ^ = A t , a A χ , a sinh χ A θ , a sinh χ sin θ A ϕ A^( hat(mu))=(A^(t),aA^(chi),a sinh chiA^(theta),a sinh chi sin thetaA^(phi))A^{\hat{\mu}}=\left(A^{t}, a A^{\chi}, a \sinh \chi A^{\theta}, a \sinh \chi \sin \theta A^{\phi}\right)Aμ^=(At,aAχ,asinhχAθ,asinhχsinθAϕ)
(b) We find
G χ ^ ϕ ^ θ ^ = G χ ϕ θ ( e θ ) θ ^ ( e χ ^ ) χ ( e ϕ ^ ) ϕ = G χ ϕ θ a sin θ . (E.89) G χ ^ ϕ ^ θ ^ = G χ ϕ θ e θ θ ^ e χ ^ χ e ϕ ^ ϕ = G χ ϕ θ a sin θ .  (E.89)  G_( hat(chi) hat(phi))^( hat(theta))=G_(chi phi)^(theta)(e_(theta))^( hat(theta))(e_( hat(chi)))^(chi)(e_( hat(phi)))^(phi)=(G_(chi phi)^(theta))/(a sin theta).quad" (E.89) "G_{\hat{\chi} \hat{\phi}}^{\hat{\theta}}=G_{\chi \phi}^{\theta}\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}\left(\boldsymbol{e}_{\hat{\chi}}\right)^{\chi}\left(\boldsymbol{e}_{\hat{\phi}}\right)^{\phi}=\frac{G_{\chi \phi}^{\theta}}{a \sin \theta} . \quad \text { (E.89) }Gχ^ϕ^θ^=Gχϕθ(eθ)θ^(eχ^)χ(eϕ^)ϕ=Gχϕθasinθ. (E.89) 
(10.2) (a) We find Γ θ ^ θ ^ θ ^ = 1 / r , Γ θ ^ r ^ θ ^ = 1 / r Γ θ ^ θ ^ θ ^ = 1 / r , Γ θ ^ r ^ θ ^ = 1 / r Gamma_( hat(theta) hat(theta))^( hat(theta))=-1//r,Gamma_( hat(theta) hat(r))^( hat(theta))=1//r\Gamma_{\hat{\theta} \hat{\theta}}^{\hat{\theta}}=-1 / r, \Gamma_{\hat{\theta} \hat{r}}^{\hat{\theta}}=1 / rΓθ^θ^θ^=1/r,Γθ^r^θ^=1/r and Γ θ ^ θ ^ θ ^ = 0 Γ θ ^ θ ^ θ ^ = 0 Gamma_( hat(theta) hat(theta))^( hat(theta))=0\Gamma_{\hat{\theta} \hat{\theta}}^{\hat{\theta}}=0Γθ^θ^θ^=0. (b) From Example 3.3, we have that [ e r ^ , e θ ^ ] = r ^ θ ^ e θ ^ / r e r ^ , e θ ^ = r ^ θ ^ e θ ^ / r [e_( hat(r)),e_( hat(theta))]=^( hat(r) hat(theta))-e_( hat(theta))//r\left[\boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}\right] \stackrel{\hat{r} \hat{\theta}}{=}-\boldsymbol{e}_{\hat{\theta}} / r[er^,eθ^]=r^θ^eθ^/r, and so ω θ ^ , [ e r ^ , e θ ^ ] = 1 / r ω θ ^ , e r ^ , e θ ^ = 1 / r (:omega^( hat(theta)),[e_( hat(r)),e_( hat(theta))]:)=-1//r\left\langle\boldsymbol{\omega}^{\hat{\theta}},\left[\boldsymbol{e}_{\hat{r}}, \boldsymbol{e}_{\hat{\theta}}\right]\right\rangle=-1 / rωθ^,[er^,eθ^]=1/r, showing that the rule is obeyed for Γ r ^ ^ θ ^ Γ θ ^ θ ^ r ^ Γ r ^ ^ θ ^ Γ θ ^ θ ^ r ^ Gamma_( hat(hat(r)) hat(theta))-Gamma^( hat(theta))_( hat(theta) hat(r))\Gamma_{\hat{\hat{r}} \hat{\theta}}-\Gamma^{\hat{\theta}}{ }_{\hat{\theta} \hat{r}}Γr^^θ^Γθ^θ^r^.
(10.3) (a) Use u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1 to say
(E.90) g μ μ ( u μ ) 2 = 1 (E.90) g μ μ u μ 2 = 1 {:(E.90)g_(mu mu)(u^(mu))^(2)=-1:}\begin{equation*} g_{\mu \mu}\left(u^{\mu}\right)^{2}=-1 \tag{E.90} \end{equation*}(E.90)gμμ(uμ)2=1
The particle doesn't move in the θ θ theta\thetaθ and ϕ ϕ phi\phiϕ directions, so we have u t = g t t a = a / g t t u t = g t t a = a / g t t u^(t)=g^(tt)a=a//g_(tt)u^{t}=g^{t t} a=a / g_{t t}ut=gtta=a/gtt
(E.91) a 2 / g t t + g r r ( u r ) 2 = 1 (E.91) a 2 / g t t + g r r u r 2 = 1 {:(E.91)a^(2)//g_(tt)+g_(rr)(u^(r))^(2)=-1:}\begin{equation*} a^{2} / g_{t t}+g_{r r}\left(u^{r}\right)^{2}=-1 \tag{E.91} \end{equation*}(E.91)a2/gtt+grr(ur)2=1
and so
(E.92) u r = [ a 2 g t t g t t g r r ] 1 2 (E.92) u r = a 2 g t t g t t g r r 1 2 {:(E.92)u^(r)=[(-a^(2)-g_(tt))/(g_(tt)g_(rr))]^((1)/(2)):}\begin{equation*} u^{r}=\left[\frac{-a^{2}-g_{t t}}{g_{t t} g_{r r}}\right]^{\frac{1}{2}} \tag{E.92} \end{equation*}(E.92)ur=[a2gttgttgrr]12
(b) Since u = d x / d τ u = d x / d τ u=dx//dtau\boldsymbol{u}=\mathrm{d} \boldsymbol{x} / \mathrm{d} \tauu=dx/dτ, write
(E.93) d r d t = d r d τ d τ d t = u r u t = g t t u r a (E.93) d r d t = d r d τ d τ d t = u r u t = g t t u r a {:(E.93)(dr)/((d)t)=(dr)/((d)tau)*((d)tau)/((d)t)=(u^(r))/(u^(t))=(g_(tt)u^(r))/(a):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{\mathrm{d} r}{\mathrm{~d} \tau} \cdot \frac{\mathrm{~d} \tau}{\mathrm{~d} t}=\frac{u^{r}}{u^{t}}=\frac{g_{t t} u^{r}}{a} \tag{E.93} \end{equation*}(E.93)dr dt=dr dτ dτ dt=urut=gttura
(c) A local observer measures time intervals d t ^ = d t ^ = d hat(t)=\mathrm{d} \hat{t}=dt^= ( e t ) t d t = g t t d t e t t d t = g t t d t (e_(t))^(t)dt=sqrt(-g_(tt))dt\left(\boldsymbol{e}_{t}\right)^{t} \mathrm{~d} t=\sqrt{-g_{t t}} \mathrm{~d} t(et)t dt=gtt dt and radial intervals d r = ( e r ) r ^ d r d r = e r r ^ d r dr=(e_(r))^( hat(r))dr\mathrm{d} r=\left(\boldsymbol{e}_{r}\right)^{\hat{r}} \mathrm{~d} rdr=(er)r^ dr, so that we have
d r ^ d t ^ = ( e r ) r ^ ( e t ) t ^ d r d t = g r r g t t g t t u r a (E.94) = g r r g t t u r a = 1 + g t t / a 2 d r ^ d t ^ = e r r ^ e t t ^ d r d t = g r r g t t g t t u r a (E.94) = g r r g t t u r a = 1 + g t t / a 2 {:[(d( hat(r)))/((d)( hat(t)))=((e_(r))^( hat(r)))/((e_(t))^( hat(t)))((d)r)/((d)t)=(sqrt(g_(rr)))/(sqrt(-g_(tt)))*(g_(tt)u^(r))/(a)],[(E.94)=sqrt(-g_(rr)g_(tt))(u^(r))/(a)=sqrt(1+g_(tt)//a^(2))]:}\begin{align*} \frac{\mathrm{d} \hat{r}}{\mathrm{~d} \hat{t}} & =\frac{\left(\boldsymbol{e}_{r}\right)^{\hat{r}}}{\left(\boldsymbol{e}_{t}\right)^{\hat{t}}} \frac{\mathrm{~d} r}{\mathrm{~d} t}=\frac{\sqrt{g_{r r}}}{\sqrt{-g_{t t}}} \cdot \frac{g_{t t} u^{r}}{a} \\ & =\sqrt{-g_{r r} g_{t t}} \frac{u^{r}}{a}=\sqrt{1+g_{t t} / a^{2}} \tag{E.94} \end{align*}dr^ dt^=(er)r^(et)t^ dr dt=grrgttgttura(E.94)=grrgttura=1+gtt/a2
(10.5) Observers at both the emission and observation points are at rest in the rotating frame, so will have velocity with non-zero component u 0 u 0 u^(0)u^{0}u0 given by
(E.95) g 00 ( u 0 ) 2 = 1 (E.95) g 00 u 0 2 = 1 {:(E.95)g_(00)(u^(0))^(2)=-1:}\begin{equation*} g_{00}\left(u^{0}\right)^{2}=-1 \tag{E.95} \end{equation*}(E.95)g00(u0)2=1
or
(E.96) u 0 = ( 1 Ω 2 r 2 ) 1 2 (E.96) u 0 = 1 Ω 2 r 2 1 2 {:(E.96)u^(0)=(1-Omega^(2)r^(2))^(-(1)/(2)):}\begin{equation*} u^{0}=\left(1-\Omega^{2} r^{2}\right)^{-\frac{1}{2}} \tag{E.96} \end{equation*}(E.96)u0=(1Ω2r2)12
Since both emitter and observer are at the same value of r r rrr, then they both find the same value of E = p u E = p u E=-p*uE=-\boldsymbol{p} \cdot \boldsymbol{u}E=pu for a photon of momentum p p p\boldsymbol{p}p.
(E.97) η μ ^ ν ^ = ( 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 ) (E.97) η μ ^ ν ^ = 0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 {:(E.97)eta_( hat(mu) hat(nu))=([0,1,0,0],[1,0,0,0],[0,0,0,-1],[0,0,-1,0]):}\eta_{\hat{\mu} \hat{\nu}}=\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \tag{E.97}\\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{array}\right)(E.97)ημ^ν^=(0100100000010010)
(b) The 1 -forms are
(E.98) ω 1 ^ = n , ω 2 ^ = l ω 3 = m ¯ , ω 4 = m (E.98) ω 1 ^ = n , ω 2 ^ = l ω 3 = m ¯ , ω 4 = m {:(E.98){:[omega^( hat(1))=n",",omega^( hat(2))=l],[omega^(3)=- bar(m)",",omega^(4)=-m]:}:}\begin{array}{cc} \omega^{\hat{1}}=n, & \omega^{\hat{2}}=\boldsymbol{l} \tag{E.98}\\ \boldsymbol{\omega}^{3}=-\bar{m}, & \boldsymbol{\omega}^{4}=-\boldsymbol{m} \end{array}(E.98)ω1^=n,ω2^=lω3=m¯,ω4=m
(11.2) (a)(i) R r r t t = 2 M r 2 ( 2 M r ) R r r t t = 2 M r 2 ( 2 M r ) R_(rrt)^(t)=(2M)/(r^(2)(2M-r))R_{r r t}^{t}=\frac{2 M}{r^{2}(2 M-r)}Rrrtt=2Mr2(2Mr).
(ii) We have that R r t r t = R t r t r = g t t R t r t r R r t r t = R t r t r = g t t R t r t r R_(rtrt)=R_(trtr)=g_(tt)R^(t)_(rtr)R_{r t r t}=R_{t r t r}=g_{t t} R^{t}{ }_{r t r}Rrtrt=Rtrtr=gttRtrtr, and so
R t r t r = g r r g t t R r t r t = 2 M ( 2 M r ) r 4 R t r t r = g r r g t t R r t r t = 2 M ( 2 M r ) r 4 R_(trt)^(r)=g^(rr)g_(tt)R_(rtr)^(t)=(2M(2M-r))/(r^(4))R_{t r t}^{r}=g^{r r} g_{t t} R_{r t r}^{t}=\frac{2 M(2 M-r)}{r^{4}}Rtrtr=grrgttRrtrt=2M(2Mr)r4
(12.1) The transformation is carried out using the transformation components
Λ x r = sin θ cos ϕ , Λ x θ = r cos θ cos ϕ , Λ x = r sin θ sin ϕ , Λ y = sin θ sin ϕ , Λ θ y = r cos θ sin ϕ , Λ ϕ y = r sin θ cos ϕ Λ r z = cos θ , Λ z θ = r sin θ Λ z ϕ = 0 Λ x r = sin θ cos ϕ , Λ x θ = r cos θ cos ϕ , Λ x = r sin θ sin ϕ , Λ y = sin θ sin ϕ , Λ θ y = r cos θ sin ϕ , Λ ϕ y = r sin θ cos ϕ Λ r z = cos θ , Λ z θ = r sin θ Λ z ϕ = 0 {:[Lambda^(x)_(r)=sin theta cos phi",",Lambda^(x)_(theta)=r cos theta cos phi","],[Lambda^(x)=-r sin theta sin phi",",Lambda^(y)=sin theta sin phi","],[Lambda_(theta)^(y)=r cos theta sin phi",",Lambda_(phi)^(y)=r sin theta cos phi],[Lambda_(r)^(z)=cos theta",",Lambda^(z)_(theta)=-r sin theta],[Lambda^(z)_(phi)=0]:}\begin{array}{cc} \Lambda^{x}{ }_{r}=\sin \theta \cos \phi, & \Lambda^{x}{ }_{\theta}=r \cos \theta \cos \phi, \\ \Lambda^{x}=-r \sin \theta \sin \phi, & \Lambda^{y}=\sin \theta \sin \phi, \\ \Lambda_{\theta}^{y}=r \cos \theta \sin \phi, & \Lambda_{\phi}^{y}=r \sin \theta \cos \phi \\ \Lambda_{r}^{z}=\cos \theta, & \Lambda^{z}{ }_{\theta}=-r \sin \theta \\ \Lambda^{z}{ }_{\phi}=0 \end{array}Λxr=sinθcosϕ,Λxθ=rcosθcosϕ,Λx=rsinθsinϕ,Λy=sinθsinϕ,Λθy=rcosθsinϕ,Λϕy=rsinθcosϕΛrz=cosθ,Λzθ=rsinθΛzϕ=0
Using these we find
T t t = ρ , T r r = p , T θ θ = p r 2 , T ϕ ϕ = p r 2 sin 2 θ T t t = ρ , T r r = p , T θ θ = p r 2 , T ϕ ϕ = p r 2 sin 2 θ T_(tt)=rho,quadT_(rr)=p,quadT_(theta theta)=pr^(2),quadT_(phi phi)=pr^(2)sin^(2)thetaT_{t t}=\rho, \quad T_{r r}=p, \quad T_{\theta \theta}=p r^{2}, \quad T_{\phi \phi}=p r^{2} \sin ^{2} \thetaTtt=ρ,Trr=p,Tθθ=pr2,Tϕϕ=pr2sin2θ.
(12.3) To apply the rule given in the question, write T μ ν ( x ) = n d τ m z ˙ n μ z ˙ n ν δ ( 3 ) [ x z n ( τ ) ] δ [ x 0 z n 0 ( τ ) ] T μ ν ( x ) = n d τ m z ˙ n μ z ˙ n ν δ ( 3 ) x z n ( τ ) δ x 0 z n 0 ( τ ) T^(mu nu)(x)=sum_(n)intdtau mz^(˙)_(n)^(mu)z^(˙)_(n)^(nu)delta^((3))[x-z_(n)(tau)]delta[x^(0)-z_(n)^(0)(tau)]T^{\mu \nu}(x)=\sum_{n} \int \mathrm{~d} \tau m \dot{z}_{n}^{\mu} \dot{z}_{n}^{\nu} \delta^{(3)}\left[x-z_{n}(\tau)\right] \delta\left[x^{0}-z_{n}^{0}(\tau)\right]Tμν(x)=n dτmz˙nμz˙nνδ(3)[xzn(τ)]δ[x0zn0(τ)], (E.101) then take f ( τ n ) = x 0 z n 0 ( τ n ) f τ n = x 0 z n 0 τ n f(tau_(n))=x^(0)-z_(n)^(0)(tau_(n))f\left(\tau_{n}\right)=x^{0}-z_{n}^{0}\left(\tau_{n}\right)f(τn)=x0zn0(τn), with zeros at τ n τ n tau_(n)\tau_{n}τn when x 0 = z 0 ( τ n ) x 0 = z 0 τ n x^(0)=z^(0)(tau_(n))x^{0}=z^{0}\left(\tau_{n}\right)x0=z0(τn) and derivative | f ( τ n ) | = | d z n 0 ( τ n ) / d τ n | f τ n = d z n 0 τ n / d τ n |f^(')(tau_(n))|=|dz_(n)^(0)(tau_(n))//dtau_(n)|\left|f^{\prime}\left(\tau_{n}\right)\right|=\left|\mathrm{d} z_{n}^{0}\left(\tau_{n}\right) / \mathrm{d} \tau_{n}\right||f(τn)|=|dzn0(τn)/dτn|. Doing the integral, we obtain an energy-momentum tensor
T μ ν ( x ) = n m z ˙ n μ ( τ n ) z ˙ n ν ( τ n ) | d z n 0 d τ n | δ ( 3 ) [ x z n ( τ n ) ] T μ ν ( x ) = n m z ˙ n μ τ n z ˙ n ν τ n d z n 0 d τ n δ ( 3 ) x z n τ n T^(mu nu)(x)=sum_(n)m(z^(˙)_(n)^(mu)(tau_(n))z^(˙)_(n)^(nu)(tau_(n)))/(|(dz_(n)^(0))/((d)tau_(n))|)delta^((3))[x-z_(n)(tau_(n))]T^{\mu \nu}(x)=\sum_{n} m \frac{\dot{z}_{n}^{\mu}\left(\tau_{n}\right) \dot{z}_{n}^{\nu}\left(\tau_{n}\right)}{\left|\frac{\mathrm{d} z_{n}^{0}}{\mathrm{~d} \tau_{n}}\right|} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right]Tμν(x)=nmz˙nμ(τn)z˙nν(τn)|dzn0 dτn|δ(3)[xzn(τn)]. (E.102)
(a) Set μ = α μ = α mu=alpha\mu=\alphaμ=α and ν = 0 ν = 0 nu=0\nu=0ν=0 and we have
T α 0 ( x ) = n m z ˙ n α ( τ n ) z ˙ n 0 ( τ n ) | z ˙ n 0 ( τ n ) | δ ( 3 ) [ x z n ( τ n ) ] = n p n α ( τ n ) δ ( 3 ) [ x z n ( τ n ) ] T α 0 ( x ) = n m z ˙ n α τ n z ˙ n 0 τ n z ˙ n 0 τ n δ ( 3 ) x z n τ n = n p n α τ n δ ( 3 ) x z n τ n {:[T^(alpha0)(x)=sum_(n)m(z^(˙)_(n)^(alpha)(tau_(n))z^(˙)_(n)^(0)(tau_(n)))/(|z^(˙)_(n)^(0)(tau_(n))|)delta^((3))[x-z_(n)(tau_(n))]],[=sum_(n)p_(n)^(alpha)(tau_(n))delta^((3))[x-z_(n)(tau_(n))]]:}\begin{aligned} T^{\alpha 0}(x) & =\sum_{n} m \frac{\dot{z}_{n}^{\alpha}\left(\tau_{n}\right) \dot{z}_{n}^{0}\left(\tau_{n}\right)}{\left|\dot{z}_{n}^{0}\left(\tau_{n}\right)\right|} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \\ & =\sum_{n} p_{n}^{\alpha}\left(\tau_{n}\right) \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \end{aligned}Tα0(x)=nmz˙nα(τn)z˙n0(τn)|z˙n0(τn)|δ(3)[xzn(τn)]=npnα(τn)δ(3)[xzn(τn)]
where we've written p n μ ( τ n ) = m z ˙ n μ ( τ n ) p n μ τ n = m z ˙ n μ τ n p_(n)^(mu)(tau_(n))=mz^(˙)_(n)^(mu)(tau_(n))p_{n}^{\mu}\left(\tau_{n}\right)=m \dot{z}_{n}^{\mu}\left(\tau_{n}\right)pnμ(τn)=mz˙nμ(τn).
(b) Set μ = α μ = α mu=alpha\mu=\alphaμ=α and ν = i ν = i nu=i\nu=iν=i and we have
T α i = n m z ˙ n α ( τ n ) z ˙ n i ( τ n ) | z ˙ n 0 ( τ n ) | δ ( 3 ) [ x z n ( τ n ) ] (E.104) = n p n α ( τ n ) δ ( 3 ) [ x z n ( τ n ) ] d z i ( τ n ) d τ n | τ n z 0 ( τ n ) | T α i = n m z ˙ n α τ n z ˙ n i τ n z ˙ n 0 τ n δ ( 3 ) x z n τ n (E.104) = n p n α τ n δ ( 3 ) x z n τ n d z i τ n d τ n τ n z 0 τ n {:[T^(alpha i)=sum_(n)m(z^(˙)_(n)^(alpha)(tau_(n))z^(˙)_(n)^(i)(tau_(n)))/(|z^(˙)_(n)^(0)(tau_(n))|)delta^((3))[x-z_(n)(tau_(n))]],[(E.104)=sum_(n)p_(n)^(alpha)(tau_(n))delta^((3))[x-z_(n)(tau_(n))](dz^(i)(tau_(n)))/(dtau_(n))|(deltau_(n))/(delz^(0)(tau_(n)))|]:}\begin{align*} T^{\alpha i} & =\sum_{n} m \frac{\dot{z}_{n}^{\alpha}\left(\tau_{n}\right) \dot{z}_{n}^{i}\left(\tau_{n}\right)}{\left|\dot{z}_{n}^{0}\left(\tau_{n}\right)\right|} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \\ & =\sum_{n} p_{n}^{\alpha}\left(\tau_{n}\right) \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right] \frac{\mathrm{d} z^{i}\left(\tau_{n}\right)}{\mathrm{d} \tau_{n}}\left|\frac{\partial \tau_{n}}{\partial z^{0}\left(\tau_{n}\right)}\right| \tag{E.104} \end{align*}Tαi=nmz˙nα(τn)z˙ni(τn)|z˙n0(τn)|δ(3)[xzn(τn)](E.104)=npnα(τn)δ(3)[xzn(τn)]dzi(τn)dτn|τnz0(τn)|
and the answer follows from taking z 0 ( τ n ) = t z 0 τ n = t z^(0)(tau_(n))=tz^{0}\left(\tau_{n}\right)=tz0(τn)=t.
(c) Use the fact that m d z n 0 / d τ n m d z n 0 / d τ n mdz_(n)^(0)//dtau_(n)m \mathrm{~d} z_{n}^{0} / \mathrm{d} \tau_{n}m dzn0/dτn is equal to the energy E n ( τ n ) E n τ n E_(n)(tau_(n))E_{n}\left(\tau_{n}\right)En(τn) to write
T α β = n m 2 z ˙ n α ( τ n ) z ˙ n β ( τ n ) E n ( τ n ) δ ( 3 ) [ x z n ( τ n ) ] , (E.105) T α β = n m 2 z ˙ n α τ n z ˙ n β τ n E n τ n δ ( 3 ) x z n τ n ,  (E.105)  T^(alpha beta)=sum_(n)m^(2)(z^(˙)_(n)^(alpha)(tau_(n))z^(˙)_(n)^(beta)(tau_(n)))/(E_(n)(tau_(n)))delta^((3))[x-z_(n)(tau_(n))],quad" (E.105) "T^{\alpha \beta}=\sum_{n} m^{2} \frac{\dot{z}_{n}^{\alpha}\left(\tau_{n}\right) \dot{z}_{n}^{\beta}\left(\tau_{n}\right)}{E_{n}\left(\tau_{n}\right)} \delta^{(3)}\left[x-z_{n}\left(\tau_{n}\right)\right], \quad \text { (E.105) }Tαβ=nm2z˙nα(τn)z˙nβ(τn)En(τn)δ(3)[xzn(τn)], (E.105) 
then rewrite in terms of momentum.
(12.4) The rule we have to apply is
(E.106) 0 = T , μ μ ν + Γ μ α μ T α ν + Γ μ α ν T α μ (E.106) 0 = T , μ μ ν + Γ μ α μ T α ν + Γ μ α ν T α μ {:(E.106)0=T_(,mu)^(mu nu)+Gamma_(mu alpha)^(mu)T^(alpha nu)+Gamma_(mu alpha)^(nu)T^(alpha mu):}\begin{equation*} 0=T_{, \mu}^{\mu \nu}+\Gamma_{\mu \alpha}^{\mu} T^{\alpha \nu}+\Gamma_{\mu \alpha}^{\nu} T^{\alpha \mu} \tag{E.106} \end{equation*}(E.106)0=T,μμν+ΓμαμTαν+ΓμανTαμ
The comma demands a derivative with respect to x μ x μ x^(mu)x^{\mu}xμ and not z μ z μ z^(mu)z^{\mu}zμ, so on contracting on the μ μ mu\muμ index, we obtain
T , μ μ ν = m ( g ) 1 2 x μ z ˙ μ z ˙ ν δ ( 4 ) [ x z ( τ ) ] d τ + m g z ˙ μ z ˙ ν x μ δ ( 4 ) [ x z ( τ ) ] d τ . (E.107) T , μ μ ν = m ( g ) 1 2 x μ z ˙ μ z ˙ ν δ ( 4 ) [ x z ( τ ) ] d τ + m g z ˙ μ z ˙ ν x μ δ ( 4 ) [ x z ( τ ) ] d τ .  (E.107)  {:[T_(,mu)^(mu nu)=m(del(-g)^(-(1)/(2)))/(delx^(mu))intz^(˙)^(mu)z^(˙)^(nu)delta^((4))[x-z(tau)]dtau],[+(m)/(sqrtg)intz^(˙)^(mu)z^(˙)^(nu)(del)/(delx^(mu))*delta^((4))[x-z(tau)]dtau.quad" (E.107) "]:}\begin{aligned} T_{, \mu}^{\mu \nu}= & m \frac{\partial(-g)^{-\frac{1}{2}}}{\partial x^{\mu}} \int \dot{z}^{\mu} \dot{z}^{\nu} \delta^{(4)}[x-z(\tau)] \mathrm{d} \tau \\ & +\frac{m}{\sqrt{g}} \int \dot{z}^{\mu} \dot{z}^{\nu} \frac{\partial}{\partial x^{\mu}} \cdot \delta^{(4)}[x-z(\tau)] \mathrm{d} \tau . \quad \text { (E.107) } \end{aligned}T,μμν=m(g)12xμz˙μz˙νδ(4)[xz(τ)]dτ+mgz˙μz˙νxμδ(4)[xz(τ)]dτ. (E.107) 
Use the identity 1 g g x μ = Γ α μ α 1 g g x μ = Γ α μ α (1)/(sqrt(-g))(delsqrt(-g))/(delx^(mu))=Gamma^(alpha)_(mu alpha)\frac{1}{\sqrt{-g}} \frac{\partial \sqrt{-g}}{\partial x^{\mu}}=\Gamma^{\alpha}{ }_{\mu \alpha}1ggxμ=Γαμα, to find that the second term becomes Γ α μ α T μ ν Γ α μ α T μ ν -Gamma^(alpha)_(mu alpha)T^(mu nu)-\Gamma^{\alpha}{ }_{\mu \alpha} T^{\mu \nu}ΓαμαTμν which, on relabelling indices, will cancel against the second term in eqn E.106. Since the delta function depends only on the separaSince the delta function depends only on the separa-
tion of x x xxx and z z zzz, we are allowed to replace / x μ / x μ del//delx^(mu)\partial / \partial x^{\mu}/xμ with tion of x x xxx and z z zzz, we are allowed to replace / x μ / x μ del//delx^(mu)\partial / \partial x^{\mu}/xμ with
/ z μ / z μ -del//delz^(mu)-\partial / \partial z^{\mu}/zμ in the first term. We also note that the chain / z μ / z μ -del//delz^(mu)-\partial / \partial z^{\mu}/zμ in the first term
rule allows us to replace
(E.108) z ˙ μ z μ = d d τ (E.108) z ˙ μ z μ = d d τ {:(E.108)z^(˙)^(mu)(del)/(delz^(mu))=(d)/((d)tau):}\begin{equation*} \dot{z}^{\mu} \frac{\partial}{\partial z^{\mu}}=\frac{\mathrm{d}}{\mathrm{~d} \tau} \tag{E.108} \end{equation*}(E.108)z˙μzμ=d dτ
leading to
T ; μ μ ν = m g z ˙ ν d d τ δ ( 4 ) [ x z ( τ ) ] d τ + Γ ν μ α m g z ˙ μ z ˙ α δ ( 4 ) [ x z ( τ ) ] d τ T ; μ μ ν = m g z ˙ ν d d τ δ ( 4 ) [ x z ( τ ) ] d τ + Γ ν μ α m g z ˙ μ z ˙ α δ ( 4 ) [ x z ( τ ) ] d τ {:[T_(;mu)^(mu nu)=-(m)/(sqrt(-g))intz^(˙)^(nu)(d)/((d)tau)delta^((4))[x-z(tau)]dtau],[+Gamma^(nu)_(mu alpha)(m)/(sqrt(-g))intz^(˙)^(mu)z^(˙)^(alpha)delta^((4))[x-z(tau)]dtau]:}\begin{aligned} T_{; \mu}^{\mu \nu}= & -\frac{m}{\sqrt{-g}} \int \dot{z}^{\nu} \frac{\mathrm{d}}{\mathrm{~d} \tau} \delta^{(4)}[x-z(\tau)] \mathrm{d} \tau \\ & +\Gamma^{\nu}{ }_{\mu \alpha} \frac{m}{\sqrt{-g}} \int \dot{z}^{\mu} \dot{z}^{\alpha} \delta^{(4)}[x-z(\tau)] \mathrm{d} \tau \end{aligned}T;μμν=mgz˙νd dτδ(4)[xz(τ)]dτ+Γνμαmgz˙μz˙αδ(4)[xz(τ)]dτ
(E.109)
Integrating the first term by parts and assuming that the velocity field dies at infinity, we find
(E.110) ( z ¨ μ + Γ α ν μ z ˙ α z ˙ ν ) δ ( 4 ) [ x z ( τ ) ] d τ = 0 (E.110) z ¨ μ + Γ α ν μ z ˙ α z ˙ ν δ ( 4 ) [ x z ( τ ) ] d τ = 0 {:(E.110)int(z^(¨)^(mu)+Gamma_(alpha nu)^(mu)z^(˙)^(alpha)z^(˙)^(nu))delta^((4))[x-z(tau)]dtau=0:}\begin{equation*} \int\left(\ddot{z}^{\mu}+\Gamma_{\alpha \nu}^{\mu} \dot{z}^{\alpha} \dot{z}^{\nu}\right) \delta^{(4)}[x-z(\tau)] \mathrm{d} \tau=0 \tag{E.110} \end{equation*}(E.110)(z¨μ+Γανμz˙αz˙ν)δ(4)[xz(τ)]dτ=0
This gives us the geodesic equation.
(12.5) (a) Insert a vector v v vvv to obtain
(E.111) T ( , v ) = p n (E.111) T ( , v ) = p n {:(E.111)T(","v)=-pn:}\begin{equation*} \boldsymbol{T}(, \boldsymbol{v})=-\boldsymbol{p} n \tag{E.111} \end{equation*}(E.111)T(,v)=pn
because J ~ ( v ) = n J ~ ( v ) = n tilde(J)(v)=-n\tilde{\boldsymbol{J}}(\boldsymbol{v})=-nJ~(v)=n. One of the eigenvectors is, therefore
T ( , v ) = ρ v T ( , v ) = ρ v T(,v)=-rho v\boldsymbol{T}(, \boldsymbol{v})=-\rho \boldsymbol{v}T(,v)=ρv
(E.112)
(b) In an orthonormal basis with v = e 0 ^ v = e 0 ^ v=e_( hat(0))v=\boldsymbol{e}_{\hat{0}}v=e0^, we have T ( , e 0 ^ ) = ρ e 0 ^ T , e 0 ^ = ρ e 0 ^ T(,e_( hat(0)))=-rhoe_( hat(0))\boldsymbol{T}\left(, \boldsymbol{e}_{\hat{0}}\right)=-\rho \boldsymbol{e}_{\hat{0}}T(,e0^)=ρe0^ and T ( , e i ^ ) = p e i ^ T , e i ^ = p e i ^ T(,e_( hat(i)))=pe_( hat(i))\boldsymbol{T}\left(, \boldsymbol{e}_{\hat{i}}\right)=p \boldsymbol{e}_{\hat{i}}T(,ei^)=pei^. As a consequence
(E.113) T ( , X ) = X 0 ^ ρ e 0 ^ + X i ^ p e i ^ . (E.113) T ( , X ) = X 0 ^ ρ e 0 ^ + X i ^ p e i ^ . {:(E.113)T(","X)=-X^( hat(0))rhoe_( hat(0))+X^( hat(i))pe_( hat(i)).:}\begin{equation*} \boldsymbol{T}(, \boldsymbol{X})=-X^{\hat{0}} \rho \boldsymbol{e}_{\hat{0}}+X^{\hat{i}} p \boldsymbol{e}_{\hat{i}} . \tag{E.113} \end{equation*}(E.113)T(,X)=X0^ρe0^+Xi^pei^.
As a result of part (a), when we input at 1-form Y Y ¯ bar(Y)\overline{\boldsymbol{Y}}Y we have
T ( Y ~ , X ) = X 0 ^ Y 0 ^ ρ + X i ^ Y i ^ p = X 0 ^ Y 0 ^ ρ + X i ^ Y i ^ p X 0 ^ Y 0 ^ p + X 0 ^ Y 0 ^ p = X 0 ^ Y 0 ^ ( ρ + p ) + X μ ^ Y μ ^ p . E.114) T ( Y ~ , X ) = X 0 ^ Y 0 ^ ρ + X i ^ Y i ^ p = X 0 ^ Y 0 ^ ρ + X i ^ Y i ^ p X 0 ^ Y 0 ^ p + X 0 ^ Y 0 ^ p = X 0 ^ Y 0 ^ ( ρ + p ) + X μ ^ Y μ ^ p .  E.114)  {:[T( tilde(Y)","X)=-X^( hat(0))Y_( hat(0))rho+X^( hat(i))Y_( hat(i))p],[=-X^( hat(0))Y_( hat(0))rho+X^( hat(i))Y_( hat(i))p-X^( hat(0))Y_( hat(0))p+X^( hat(0))Y_( hat(0))p],[=-X^( hat(0))Y_( hat(0))(rho+p)+X^( hat(mu))Y_( hat(mu))p.quad" E.114) "]:}\begin{aligned} \boldsymbol{T}(\tilde{\boldsymbol{Y}}, \boldsymbol{X}) & =-X^{\hat{0}} Y_{\hat{0}} \rho+X^{\hat{i}} Y_{\hat{i}} p \\ & =-X^{\hat{0}} Y_{\hat{0}} \rho+X^{\hat{i}} Y_{\hat{i}} p-X^{\hat{0}} Y_{\hat{0}} p+X^{\hat{0}} Y_{\hat{0}} p \\ & =-X^{\hat{0}} Y_{\hat{0}}(\rho+p)+X^{\hat{\mu}} Y_{\hat{\mu}} p . \quad \text { E.114) } \end{aligned}T(Y~,X)=X0^Y0^ρ+Xi^Yi^p=X0^Y0^ρ+Xi^Yi^pX0^Y0^p+X0^Y0^p=X0^Y0^(ρ+p)+Xμ^Yμ^p. E.114) 
(c) The right-hand side of the last equation is a number that can be rewritten as X 0 ^ Y 0 ^ ( ρ + p ) + η μ ^ ν ^ X μ ^ Y μ ^ p X 0 ^ Y 0 ^ ( ρ + p ) + η μ ^ ν ^ X μ ^ Y μ ^ p X^( hat(0))Y^( hat(0))(rho+p)+eta_( hat(mu) hat(nu))X^( hat(mu))Y^( hat(mu))pX^{\hat{0}} Y^{\hat{0}}(\rho+p)+\eta_{\hat{\mu} \hat{\nu}} X^{\hat{\mu}} Y^{\hat{\mu}} pX0^Y0^(ρ+p)+ημ^ν^Xμ^Yμ^p, since Y 0 ^ = Y 0 Y 0 ^ = Y 0 Y_( hat(0))=-Y^(0)Y_{\hat{0}}=-Y^{0}Y0^=Y0 in flat space. Hence we can extract a ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) version of the tensor components. Using the fact that v = e 0 ^ v = e 0 ^ v=e_( hat(0))\boldsymbol{v}=\boldsymbol{e}_{\hat{0}}v=e0^, we have that v ~ v ~ tilde(v)\tilde{\boldsymbol{v}}v~ can be used to extract the timelike components of vectors. The corresponding flattimelike components of vectors. The corresponding flat-
space expression for T ( Y , X ) T ( Y , X ) T(Y,X)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X})T(Y,X) can then be reexpressed space expression for T ( Y , X ) T ( Y , X ) T(Y,X)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X})T(Y,X) can then be reexp
in vector notation using v ~ v ~ tilde(v)\tilde{\boldsymbol{v}}v~ and the metric η η eta\boldsymbol{\eta}η as
T ( Y , X ) = ( ρ + p ) [ v ~ ( Y ) v ~ ( X ) ] + p η ( Y , X ) . ( E .115 ) T ( Y , X ) = ( ρ + p ) [ v ~ ( Y ) v ~ ( X ) ] + p η ( Y , X ) . ( E .115 ) T(Y,X)=(rho+p)[ tilde(v)(Y) tilde(v)(X)]+p eta(Y,X).(E.115)\boldsymbol{T}(\boldsymbol{Y}, \boldsymbol{X})=(\rho+p)[\tilde{\boldsymbol{v}}(\boldsymbol{Y}) \tilde{\boldsymbol{v}}(\boldsymbol{X})]+p \boldsymbol{\eta}(\boldsymbol{Y}, \boldsymbol{X}) .(\mathrm{E} .115)T(Y,X)=(ρ+p)[v~(Y)v~(X)]+pη(Y,X).(E.115)
In curved space, we swap η g η g eta rarr g\boldsymbol{\eta} \rightarrow \boldsymbol{g}ηg. We therefore suggest that the form of the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor is
(E.116) T ( , ) = ( ρ + p ) v ~ v ~ + p g (E.116) T ( , ) = ( ρ + p ) v ~ v ~ + p g {:(E.116)T(",")=(rho+p) tilde(v)ox tilde(v)+pg:}\begin{equation*} \boldsymbol{T}(,)=(\rho+p) \tilde{\boldsymbol{v}} \otimes \tilde{\boldsymbol{v}}+p \boldsymbol{g} \tag{E.116} \end{equation*}(E.116)T(,)=(ρ+p)v~v~+pg
This tensor has components
(E.117) T μ ν = ( ρ + p ) v μ v ν + p g μ ν (E.117) T μ ν = ( ρ + p ) v μ v ν + p g μ ν {:(E.117)T_(mu nu)=(rho+p)v_(mu)v_(nu)+pg_(mu nu):}\begin{equation*} T_{\mu \nu}=(\rho+p) v_{\mu} v_{\nu}+p g_{\mu \nu} \tag{E.117} \end{equation*}(E.117)Tμν=(ρ+p)vμvν+pgμν
(15.1) Volume of a cylinder is π r 2 h 10 11 π r 2 h 10 11 pir^(2)h~~10^(11)\pi r^{2} h \approx 10^{11}πr2h1011 in cubic parsecs, which is then the number of stars.
(15.2) (a) The invariant interval can be written in terms of the metric line element as
s = d τ { ( d t d τ ) 2 a ( t ) 2 [ i = 1 3 ( d x i d τ ) 2 ] } ( E .118 ) 1 2 s = d τ d t d τ 2 a ( t ) 2 i = 1 3 d x i d τ 2 ( E .118 ) 1 2 s=intdtau{(((d)t)/((d)tau))^(2)-a(t)^(2)[sum_(i=1)^(3)(((d)x^(i))/((d)tau))^(2)]}_((E.118))^((1)/(2))s=\int \mathrm{d} \tau\left\{\left(\frac{\mathrm{~d} t}{\mathrm{~d} \tau}\right)^{2}-a(t)^{2}\left[\sum_{i=1}^{3}\left(\frac{\mathrm{~d} x^{i}}{\mathrm{~d} \tau}\right)^{2}\right]\right\}_{(\mathrm{E} .118)}^{\frac{1}{2}}s=dτ{( dt dτ)2a(t)2[i=13( dxi dτ)2]}(E.118)12
where the interval has been parametrized using the proper time. We find
(E.119) L t = i L , L x ˙ 1 = a 2 x ˙ i L L t = a a d t ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) L , L x i = 0 (E.119) L t = i L , L x ˙ 1 = a 2 x ˙ i L L t = a a d t x ˙ 2 + y ˙ 2 + z ˙ 2 L , L x i = 0 {:(E.119){:[(del L)/(del t)=(i)/(L)",",(del L)/(delx^(˙)^(1))=-(a^(2)x^(˙)^(i))/(L)],[(del L)/(del t)=-(a(a)/(dt)(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2)))/(L)",",(del L)/(delx^(i))=0]:}:}\begin{array}{cc} \frac{\partial L}{\partial t}=\frac{i}{L}, & \frac{\partial L}{\partial \dot{x}^{1}}=-\frac{a^{2} \dot{x}^{i}}{L} \tag{E.119}\\ \frac{\partial L}{\partial t}=-\frac{a \frac{a}{d t}\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right)}{L}, & \frac{\partial L}{\partial x^{i}}=0 \end{array}(E.119)Lt=iL,Lx˙1=a2x˙iLLt=aadt(x˙2+y˙2+z˙2)L,Lxi=0
where dots here indicate derivatives with respect to the proper time. This gives us equations of motion
x ¨ i + 2 a d a d t x ˙ i t ˙ = 0 (E.120) t ¨ + a d a d t ( x ˙ 2 + y ˙ 2 + z ˙ 2 ) = 0 x ¨ i + 2 a d a d t x ˙ i t ˙ = 0 (E.120) t ¨ + a d a d t x ˙ 2 + y ˙ 2 + z ˙ 2 = 0 {:[x^(¨)^(i)+(2)/(a)((d)a)/((d)t)*x^(˙)^(i)t^(˙)=0],[(E.120)t^(¨)+a((d)a)/((d)t)*(x^(˙)^(2)+y^(˙)^(2)+z^(˙)^(2))=0]:}\begin{align*} \ddot{x}^{i}+\frac{2}{a} \frac{\mathrm{~d} a}{\mathrm{~d} t} \cdot \dot{x}^{i} \dot{t} & =0 \\ \ddot{t}+a \frac{\mathrm{~d} a}{\mathrm{~d} t} \cdot\left(\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}\right) & =0 \tag{E.120} \end{align*}x¨i+2a da dtx˙it˙=0(E.120)t¨+a da dt(x˙2+y˙2+z˙2)=0
The connection coefficients can then be read off.
(b) The only derivatives that survive are
Γ i i , 0 0 = ( a ˙ 2 + a a ¨ ) , (E.121) Γ 0 i , 0 i = a ¨ a ( a ˙ a ) 2 , Γ i i , 0 0 = a ˙ 2 + a a ¨ , (E.121) Γ 0 i , 0 i = a ¨ a a ˙ a 2 , {:[Gamma_(ii,0)^(0)=(a^(˙)^(2)+a(a^(¨)))","],[(E.121)Gamma_(0i,0)^(i)=((a^(¨)))/(a)-(((a^(˙)))/(a))^(2)","]:}\begin{align*} \Gamma_{i i, 0}^{0} & =\left(\dot{a}^{2}+a \ddot{a}\right), \\ \Gamma_{0 i, 0}^{i} & =\frac{\ddot{a}}{a}-\left(\frac{\dot{a}}{a}\right)^{2}, \tag{E.121} \end{align*}Γii,00=(a˙2+aa¨),(E.121)Γ0i,0i=a¨a(a˙a)2,
using dots here for derivatives with respect to t t ttt. Plugging into the equation for the components of R R R\boldsymbol{R}R we find the non-zero parts are
R i 0 i 0 = Γ 0 i i , 0 Γ 0 i i Γ i i 0 = a a ¨ (E.122) R j i j i = Γ i 0 i Γ 0 j j = a ˙ 2 . R i 0 i 0 = Γ 0 i i , 0 Γ 0 i i Γ i i 0 = a a ¨ (E.122) R j i j i = Γ i 0 i Γ 0 j j = a ˙ 2 . {:[R_(i0i)^(0)=Gamma^(0)_(ii,0)-Gamma^(0)_(ii)Gamma^(i)_(i0)=aa^(¨)],[(E.122)R_(jij)^(i)=Gamma^(i)_(0i)Gamma^(0)_(jj)=a^(˙)^(2).]:}\begin{align*} & R_{i 0 i}^{0}=\Gamma^{0}{ }_{i i, 0}-\Gamma^{0}{ }_{i i} \Gamma^{i}{ }_{i 0}=a \ddot{a} \\ & R_{j i j}^{i}=\Gamma^{i}{ }_{0 i} \Gamma^{0}{ }_{j j}=\dot{a}^{2} . \tag{E.122} \end{align*}Ri0i0=Γ0ii,0Γ0iiΓii0=aa¨(E.122)Rjiji=Γi0iΓ0jj=a˙2.
These can be shifted into the orthonormal frame using the vielbein given in Chapter 15.
(15.3) (a) The Einstein tensor for the state of affairs described is given by
(E.123) G μ ν = R μ ν 1 2 R g μ ν = 1 4 R g μ ν (E.123) G μ ν = R μ ν 1 2 R g μ ν = 1 4 R g μ ν {:(E.123)G_(mu nu)=R_(mu nu)-(1)/(2)Rg_(mu nu)=-(1)/(4)Rg_(mu nu):}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu}=-\frac{1}{4} R g_{\mu \nu} \tag{E.123} \end{equation*}(E.123)Gμν=Rμν12Rgμν=14Rgμν
(b) The components G μ ν G μ ν G_(mu nu)G_{\mu \nu}Gμν must be equal to the energymomentum contribution ( = Λ g μ ν ) = Λ g μ ν (=-Lambdag_(mu nu))\left(=-\Lambda g_{\mu \nu}\right)(=Λgμν), so that the Einstein equation reads
(E.124) R 4 g μ ν = Λ g μ ν (E.124) R 4 g μ ν = Λ g μ ν {:(E.124)-(R)/(4)g_(mu nu)=-Lambdag_(mu nu):}\begin{equation*} -\frac{R}{4} g_{\mu \nu}=-\Lambda g_{\mu \nu} \tag{E.124} \end{equation*}(E.124)R4gμν=Λgμν
and we have Λ = R / 4 Λ = R / 4 Lambda=R//4\Lambda=R / 4Λ=R/4.
(c) The Ricci scalar is given by
(E.125) R = 6 ( a ¨ a + a ˙ 2 a 2 ) (E.125) R = 6 a ¨ a + a ˙ 2 a 2 {:(E.125)R=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2))):}\begin{equation*} R=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right) \tag{E.125} \end{equation*}(E.125)R=6(a¨a+a˙2a2)
with a ( t ) = e H t a ( t ) = e H t a(t)=e^(Ht)a(t)=e^{H t}a(t)=eHt. Substituting, we find that
(E.126) R = 6 ( H 2 + H 2 ) = 12 Λ 3 = 4 Λ (E.126) R = 6 H 2 + H 2 = 12 Λ 3 = 4 Λ {:(E.126)R=6(H^(2)+H^(2))=12(Lambda)/(3)=4Lambda:}\begin{equation*} R=6\left(H^{2}+H^{2}\right)=12 \frac{\Lambda}{3}=4 \Lambda \tag{E.126} \end{equation*}(E.126)R=6(H2+H2)=12Λ3=4Λ
So fixing a Ricci scalar R = 4 Λ R = 4 Λ R=4LambdaR=4 \LambdaR=4Λ gives us the same flat, expanding Universe that we had before.
(15.5) (a) The energy density is of order 1 GeV / ( 10 45 m 3 ) 1 GeV / 10 45 m 3 1GeV//(10^(-45)m^(3))~~1 \mathrm{GeV} /\left(10^{-45} \mathrm{~m}^{3}\right) \approx1GeV/(1045 m3) 1.602 × 10 35 Jm 3 1.602 × 10 35 Jm 3 1.602 xx10^(35)Jm^(-3)1.602 \times 10^{35} \mathrm{Jm}^{-3}1.602×1035Jm3.
(b) In SI units, the Einstein equation is
(E.127) G = 8 π G c 4 T (E.127) G = 8 π G c 4 T {:(E.127)G=(8pi G)/(c^(4))T:}\begin{equation*} \boldsymbol{G}=\frac{8 \pi G}{c^{4}} \boldsymbol{T} \tag{E.127} \end{equation*}(E.127)G=8πGc4T
where the factor 8 π G c 4 = 2.07 × 10 43 m 1 kg 1 s 2 8 π G c 4 = 2.07 × 10 43 m 1 kg 1 s 2 (8pi G)/(c^(4))=2.07 xx10^(-43)m^(-1)kg^(-1)s^(2)\frac{8 \pi G}{c^{4}}=2.07 \times 10^{-43} \mathrm{~m}^{-1} \mathrm{~kg}^{-1} \mathrm{~s}^{2}8πGc4=2.07×1043 m1 kg1 s2. The right-hand-side is therefore 3 × 10 8 m 2 3 × 10 8 m 2 ~~3xx10^(-8)m^(-2)\approx 3 \times 10^{-8} \mathrm{~m}^{-2}3×108 m2. Take the left-hand side to be equal to the curvature 1 / a 2 1 / a 2 1//a^(2)1 / a^{2}1/a2, where a a aaa is the characteristic size of the Universe, we have a 5 km a 5 km a~~5kma \approx 5 \mathrm{~km}a5 km. We conclude that the vacuum energy of the strong nuclear force would wind the Universe into a ball with a radius of a few km . We could barely go for a walk before finding we are back where we started!
(15.6) (a) The equation of motion for t t ttt is
(E.128) d 2 t d τ 2 + a a ˙ ( d r d τ ) 2 = 0 (E.128) d 2 t d τ 2 + a a ˙ d r d τ 2 = 0 {:(E.128)(d^(2)t)/((d)tau^(2))+aa^(˙)(((d)r)/((d)tau))^(2)=0:}\begin{equation*} \frac{\mathrm{d}^{2} t}{\mathrm{~d} \tau^{2}}+a \dot{a}\left(\frac{\mathrm{~d} r}{\mathrm{~d} \tau}\right)^{2}=0 \tag{E.128} \end{equation*}(E.128)d2t dτ2+aa˙( dr dτ)2=0
The r r rrr-dependence can be eliminated by substituting using the length parametrization condition L = 1 L = 1 L=1L=1L=1.
(c) Substituting into the velocity identity g μ ν u μ u ν = g μ ν u μ u ν = g_(mu nu)u^(mu)u^(nu)=g_{\mu \nu} u^{\mu} u^{\nu}=gμνuμuν= -1 , we find
(E.129) ( d t d τ ) 2 + g i i ( u i ) 2 = 1 (E.129) d t d τ 2 + g i i u i 2 = 1 {:(E.129)-((dt)/((d)tau))^(2)+g_(ii)(u^(i))^(2)=-1:}\begin{equation*} -\left(\frac{\mathrm{d} t}{\mathrm{~d} \tau}\right)^{2}+g_{i i}\left(u^{i}\right)^{2}=-1 \tag{E.129} \end{equation*}(E.129)(dt dτ)2+gii(ui)2=1
If we take g i i ( u i ) 2 = | u | 2 g i i u i 2 = | u | 2 g_(ii)(u^(i))^(2)=| vec(u)|^(2)g_{i i}\left(u^{i}\right)^{2}=|\vec{u}|^{2}gii(ui)2=|u|2 then using the solution of the equation of motion to substitute for d t / d τ d t / d τ dt//dtau\mathrm{d} t / \mathrm{d} \taudt/dτ yields the answer.
(15.7) (a) Take d s = 0 d s = 0 ds=0\mathrm{d} s=0ds=0 and write d t / a ( t ) = d r d t / a ( t ) = d r dt//a(t)=dr\mathrm{d} t / a(t)=\mathrm{d} rdt/a(t)=dr, or
(E.130) 0 d r = t s t r d t e H t (E.130) 0 d r = t s t r d t e H t {:(E.130)int_(0)^(ℓ)dr=int_(t_(s))^(t_(r))dte^(-Ht):}\begin{equation*} \int_{0}^{\ell} \mathrm{d} r=\int_{t_{\mathrm{s}}}^{t_{\mathrm{r}}} \mathrm{~d} t \mathrm{e}^{-H t} \tag{E.130} \end{equation*}(E.130)0dr=tstr dteHt
from which the answer follows.
(b) As \ell increases, t r t r t_(r)t_{\mathrm{r}}tr also increases. The distance = e H t s / H = e H t s / H ℓ^(')=e^(-Ht_(s))//H\ell^{\prime}=\mathrm{e}^{-H t_{\mathrm{s}}} / H=eHts/H corresponds to an infinite receipt time t r t r t_(r)rarr oot_{\mathrm{r}} \rightarrow \inftytr. Therefore, an observer beyond a radius ℓ^(')\ell^{\prime} will never receive the light signal sent from the origin, since the expansion of the Universe is fast enough to prevent this. Notice this scale is set by 1 / H 1 / H 1//H1 / H1/H.
(15.8) From Exercise 15.2, we start from one of the equations of motion, in the form,
(E.131) d d τ ( a 2 x ˙ i ) = 0 (E.131) d d τ a 2 x ˙ i = 0 {:(E.131)(d)/((d)tau)(a^(2)x^(˙)^(i))=0:}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau}\left(a^{2} \dot{x}^{i}\right)=0 \tag{E.131} \end{equation*}(E.131)d dτ(a2x˙i)=0
This tells us that
(E.132) e 2 H t x ˙ i = const. (E.132) e 2 H t x ˙ i =  const.  {:(E.132)e^(2Ht)x^(˙)^(i)=" const. ":}\begin{equation*} \mathrm{e}^{2 H t} \dot{x}^{i}=\text { const. } \tag{E.132} \end{equation*}(E.132)e2Htx˙i= const. 
or
(E.133) x ˙ i = C i e 2 H t (E.133) x ˙ i = C i e 2 H t {:(E.133)x^(˙)^(i)=C^(i)e^(-2Ht):}\begin{equation*} \dot{x}^{i}=C^{i} \mathrm{e}^{-2 H t} \tag{E.133} \end{equation*}(E.133)x˙i=Cie2Ht
with C i C i C^(i)C^{i}Ci a set of constants. To see that the solutions are straight lines, we write
(E.134) d x i d x j = C i C j (E.134) d x i d x j = C i C j {:(E.134)(dx^(i))/((d)x^(j))=(C^(i))/(C^(j)):}\begin{equation*} \frac{\mathrm{d} x^{i}}{\mathrm{~d} x^{j}}=\frac{C^{i}}{C^{j}} \tag{E.134} \end{equation*}(E.134)dxi dxj=CiCj
This latter equation is solved with straight lines by saying, for example,
x 1 = x 0 1 + C 1 x 3 / C 3 , (E.135) x 2 = x 0 2 + C 2 x 3 / C 3 , x 1 = x 0 1 + C 1 x 3 / C 3 , (E.135) x 2 = x 0 2 + C 2 x 3 / C 3 , {:[x^(1)=x_(0)^(1)+C^(1)x^(3)//C^(3)","],[(E.135)x^(2)=x_(0)^(2)+C^(2)x^(3)//C^(3)","]:}\begin{align*} & x^{1}=x_{0}^{1}+C^{1} x^{3} / C^{3}, \\ & x^{2}=x_{0}^{2}+C^{2} x^{3} / C^{3}, \tag{E.135} \end{align*}x1=x01+C1x3/C3,(E.135)x2=x02+C2x3/C3,
with x 0 i x 0 i x_(0)^(i)x_{0}^{i}x0i another set of constants set by the initial conditions.
If computing directly from the metric elements, we obtain
Γ t r r = Γ r r t = a a ˙ 1 k r 2 , Γ t i i = Γ i i t = a ˙ a g i i , Γ r r r = a 2 k r ( 1 k r 2 ) 2 , Γ r θ θ = Γ θ θ r = a 2 r Γ r ϕ ϕ = Γ ϕ ϕ r = a 2 r sin 2 θ , Γ θ ϕ ϕ = Γ ϕ ϕ θ = a 2 r 2 sin θ cos θ . (E.136) Γ t r r = Γ r r t = a a ˙ 1 k r 2 , Γ t i i = Γ i i t = a ˙ a g i i , Γ r r r = a 2 k r 1 k r 2 2 , Γ r θ θ = Γ θ θ r = a 2 r Γ r ϕ ϕ = Γ ϕ ϕ r = a 2 r sin 2 θ , Γ θ ϕ ϕ = Γ ϕ ϕ θ = a 2 r 2 sin θ cos θ .  (E.136)  {:[Gamma_(trr)=-Gamma_(rrt)=-(a(a^(˙)))/(1-kr^(2))","],[Gamma_(tii)=-Gamma_(iit)=-((a^(˙)))/(a)g_(ii)","],[Gamma_(rrr)=(a^(2)kr)/((1-kr^(2))^(2))","],[Gamma_(r theta theta)=-Gamma_(theta theta r)=-a^(2)r],[Gamma_(r phi phi)=-Gamma_(phi phi r)=-a^(2)rsin^(2)theta","],[Gamma_(theta phi phi)=-Gamma_(phi phi theta)=-a^(2)r^(2)sin theta cos theta.quad" (E.136) "]:}\begin{aligned} \Gamma_{t r r}=-\Gamma_{r r t} & =-\frac{a \dot{a}}{1-k r^{2}}, \\ \Gamma_{t i i}=-\Gamma_{i i t} & =-\frac{\dot{a}}{a} g_{i i}, \\ \Gamma_{r r r} & =\frac{a^{2} k r}{\left(1-k r^{2}\right)^{2}}, \\ \Gamma_{r \theta \theta} & =-\Gamma_{\theta \theta r}=-a^{2} r \\ \Gamma_{r \phi \phi} & =-\Gamma_{\phi \phi r}=-a^{2} r \sin ^{2} \theta, \\ \Gamma_{\theta \phi \phi} & =-\Gamma_{\phi \phi \theta}=-a^{2} r^{2} \sin \theta \cos \theta . \quad \text { (E.136) } \end{aligned}Γtrr=Γrrt=aa˙1kr2,Γtii=Γiit=a˙agii,Γrrr=a2kr(1kr2)2,Γrθθ=Γθθr=a2rΓrϕϕ=Γϕϕr=a2rsin2θ,Γθϕϕ=Γϕϕθ=a2r2sinθcosθ. (E.136) 
The answers then follow by raising components with g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν.
(16.5) Substitute the metric into the Bianchi identity to find
C , σ ( g μ α g ν β g μ β g ν α ) + C , β ( g μ σ g ν α g μ α g ν σ ) (E.137) + C , α ( g μ β g ν σ g μ σ g ν β ) = 0 C , σ g μ α g ν β g μ β g ν α + C , β g μ σ g ν α g μ α g ν σ (E.137) + C , α g μ β g ν σ g μ σ g ν β = 0 {:[C_(,sigma)(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha))],[+C_(,beta)(g_(mu sigma)g_(nu alpha)-g_(mu alpha)g_(nu sigma))],[(E.137)+C_(,alpha)(g_(mu beta)g_(nu sigma)-g_(mu sigma)g_(nu beta))=0]:}\begin{align*} & C_{, \sigma}\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right) \\ + & C_{, \beta}\left(g_{\mu \sigma} g_{\nu \alpha}-g_{\mu \alpha} g_{\nu \sigma}\right) \\ + & C_{, \alpha}\left(g_{\mu \beta} g_{\nu \sigma}-g_{\mu \sigma} g_{\nu \beta}\right)=0 \tag{E.137} \end{align*}C,σ(gμαgνβgμβgνα)+C,β(gμσgναgμαgνσ)(E.137)+C,α(gμβgνσgμσgνβ)=0
where we've used the compatibility condition for the derivatives of the metric components (i.e. g μ ν ; σ = 0 g μ ν ; σ = 0 g_(mu nu;sigma)=0g_{\mu \nu ; \sigma}=0gμν;σ=0 ). Next, contract indices by (i) multiplying by g μ α g μ α g^(mu alpha)g^{\mu \alpha}gμα and summing, then (ii) multiplying by g ν β g ν β g^(nu beta)g^{\nu \beta}gνβ and summing. We find that C , σ = 0 C , σ = 0 C_(,sigma)=0C_{, \sigma}=0C,σ=0, so C C CCC must be constant.
(16.6) We write
(E.138) R ν β = C g μ α ( g μ α g ν β g μ β g ν α ) (E.138) R ν β = C g μ α g μ α g ν β g μ β g ν α {:(E.138)R_(nu beta)=Cg^(mu alpha)(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha)):}\begin{equation*} R_{\nu \beta}=C g^{\mu \alpha}\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right) \tag{E.138} \end{equation*}(E.138)Rνβ=Cgμα(gμαgνβgμβgνα)
from which we obtain, in (3+1) dimensions,
(E.139) R ν β = C ( 4 g ν β g ν β ) = 3 C g ν β (E.139) R ν β = C 4 g ν β g ν β = 3 C g ν β {:(E.139)R_(nu beta)=C(4g_(nu beta)-g_(nu beta))=3Cg_(nu beta):}\begin{equation*} R_{\nu \beta}=C\left(4 g_{\nu \beta}-g_{\nu \beta}\right)=3 C g_{\nu \beta} \tag{E.139} \end{equation*}(E.139)Rνβ=C(4gνβgνβ)=3Cgνβ
and so R = 12 C R = 12 C R=12 CR=12 CR=12C. This means that R ν β = 1 4 R g ν β R ν β = 1 4 R g ν β R_(nu beta)=(1)/(4)Rg_(nu beta)R_{\nu \beta}=\frac{1}{4} R g_{\nu \beta}Rνβ=14Rgνβ and also that the components of the Einstein tensor are G ν β = 1 4 R g ν β G ν β = 1 4 R g ν β G_(nu beta)=-(1)/(4)Rg_(nu beta)G_{\nu \beta}=-\frac{1}{4} R g_{\nu \beta}Gνβ=14Rgνβ. In general, we have for d d ddd-dimensional spacetime that ( d ) R μ ν = ( d 1 ) C g μ ν ( d ) R μ ν = ( d 1 ) C g μ ν ^((d))R_(mu nu)=(d-1)Cg_(mu nu){ }^{(d)} R_{\mu \nu}=(d-1) C g_{\mu \nu}(d)Rμν=(d1)Cgμν and ( d ) R = ( d ) R = ^((d))R={ }^{(d)} R=(d)R= C d ( d 1 ) C d ( d 1 ) Cd(d-1)C d(d-1)Cd(d1). This allows us to write ( d ) R μ ν = ( d ) R d g μ ν ( d ) R μ ν = ( d ) R d g μ ν ^((d))R_(mu nu)=(^((d))R)/(d)g_(mu nu){ }^{(d)} R_{\mu \nu}=\frac{{ }^{(d)} R}{d} g_{\mu \nu}(d)Rμν=(d)Rdgμν. (16.7) We have
(E.140) ( 3 ) R = 3 R i ^ i ^ = 3 a ¨ a + 6 ( a ˙ a ) 2 + 6 k a (E.140) ( 3 ) R = 3 R i ^ i ^ = 3 a ¨ a + 6 a ˙ a 2 + 6 k a {:(E.140)^((3))R=3R_( hat(i) hat(i))=3((a^(¨)))/(a)+6(((a^(˙)))/(a))^(2)+6(k)/(a):}\begin{equation*} { }^{(3)} R=3 R_{\hat{i} \hat{i}}=3 \frac{\ddot{a}}{a}+6\left(\frac{\dot{a}}{a}\right)^{2}+6 \frac{k}{a} \tag{E.140} \end{equation*}(E.140)(3)R=3Ri^i^=3a¨a+6(a˙a)2+6ka
This can be used to write ( 3 ) R i j = 1 3 ( 3 ) R γ i j ( 3 ) R i j = 1 3 ( 3 ) R γ i j ^((3))R_(ij)=(1)/(3)^((3))Rgamma_(ij){ }^{(3)} R_{i j}=\frac{1}{3}{ }^{(3)} R \gamma_{i j}(3)Rij=13(3)Rγij, in a similar fashion to the previous question.
(16.8) We have that ob / em = a ( t ob ) / a ( t em ) ob / em = a t ob / a t em ℓ_(ob)//ℓ_(em)=a(t_(ob))//a(t_(em))\ell_{\mathrm{ob}} / \ell_{\mathrm{em}}=a\left(t_{\mathrm{ob}}\right) / a\left(t_{\mathrm{em}}\right)ob/em=a(tob)/a(tem). Since a ( t ob ) / a ( t em ) = ( 1 + z ) a t ob / a t em = ( 1 + z ) a(t_(ob))//a(t_(em))=(1+z)a\left(t_{\mathrm{ob}}\right) / a\left(t_{\mathrm{em}}\right)=(1+z)a(tob)/a(tem)=(1+z), we have
(E.141) em = ob 1 + z (E.141) em = ob 1 + z {:(E.141)ℓ_(em)=(ℓ_(ob))/(1+z):}\begin{equation*} \ell_{\mathrm{em}}=\frac{\ell_{\mathrm{ob}}}{1+z} \tag{E.141} \end{equation*}(E.141)em=ob1+z
(17.1) (a) We have that
(E.142) Λ χ r = χ r = 1 ( 1 k r 2 ) 1 2 (E.142) Λ χ r = χ r = 1 1 k r 2 1 2 {:(E.142)Lambda^(chi)_(r)=(del chi)/(del r)=(1)/((1-kr^(2))^((1)/(2))):}\begin{equation*} \Lambda^{\chi}{ }_{r}=\frac{\partial \chi}{\partial r}=\frac{1}{\left(1-k r^{2}\right)^{\frac{1}{2}}} \tag{E.142} \end{equation*}(E.142)Λχr=χr=1(1kr2)12
(c) The transformation we want to check is
T r ^ r ^ = ( Λ χ ^ r ^ ) 2 T χ ^ χ ^ (E.143) = [ Λ χ r ( e χ ) χ ^ ( e r ^ ) r ] 2 T χ ^ χ ^ T r ^ r ^ = Λ χ ^ r ^ 2 T χ ^ χ ^ (E.143) = Λ χ r e χ χ ^ e r ^ r 2 T χ ^ χ ^ {:[T_( hat(r) hat(r))=(Lambda_( hat(chi))_( hat(r)))^(2)T_( hat(chi) hat(chi))],[(E.143)=[Lambda^(chi)_(r)(e_(chi))^( hat(chi))(e_( hat(r)))^(r)]^(2)T_( hat(chi) hat(chi))]:}\begin{align*} T_{\hat{r} \hat{r}} & =\left(\Lambda_{\hat{\chi}}{ }_{\hat{r}}\right)^{2} T_{\hat{\chi} \hat{\chi}} \\ & =\left[\Lambda^{\chi}{ }_{r}\left(\boldsymbol{e}_{\chi}\right)^{\hat{\chi}}\left(\boldsymbol{e}_{\hat{r}}\right)^{r}\right]^{2} T_{\hat{\chi} \hat{\chi}} \tag{E.143} \end{align*}Tr^r^=(Λχ^r^)2Tχ^χ^(E.143)=[Λχr(eχ)χ^(er^)r]2Tχ^χ^
Using the results of parts (a) and (b) we find
Λ r χ ( e χ ) χ ^ ( e r ^ ) r = 1 Λ r χ e χ χ ^ e r ^ r = 1 Lambda_(r)^(chi)(e_(chi))^( hat(chi))(e_( hat(r)))^(r)=1\Lambda_{r}^{\chi}\left(\boldsymbol{e}_{\chi}\right)^{\hat{\chi}}\left(\boldsymbol{e}_{\hat{r}}\right)^{r}=1Λrχ(eχ)χ^(er^)r=1
(17.2) (a) Using U = ( ln Z ) / β U = ( ln Z ) / β U=-del(ln Z)//del betaU=-\partial(\ln Z) / \partial \betaU=(lnZ)/β, with β = 1 / k B T β = 1 / k B T beta=1//k_(B)T\beta=1 / k_{\mathrm{B}} Tβ=1/kBT, we find that the internal energy per particle is 3 k B T 3 k B T 3k_(B)T3 k_{\mathrm{B}} T3kBT. (b) The pressure is found from p = ( F / V ) T p = ( F / V ) T p=-(del F//del V)_(T)p=-(\partial F / \partial V)_{T}p=(F/V)T, where F = 1 β ln Z F = 1 β ln Z F=-(1)/(beta)ln ZF=-\frac{1}{\beta} \ln ZF=1βlnZ is the free energy. We obtain a pressure k B T / V k B T / V k_(B)T//Vk_{\mathrm{B}} T / VkBT/V per particle. The result in (c) follows from taking the ratio of these two results.
(17.3) Setting k = 0 k = 0 k=0k=0k=0 we find
(E.144) a ˙ 2 = const (E.144) a ˙ 2 =  const  {:(E.144)a^(˙)^(2)=" const ":}\begin{equation*} \dot{a}^{2}=\text { const } \tag{E.144} \end{equation*}(E.144)a˙2= const 
so a ( t ) t a ( t ) t a(t)prop ta(t) \propto ta(t)t.
(17.4) Since k = 0 k = 0 k=0k=0k=0, we have Ω d ( 0 ) = 1 Ω d ( 0 ) = 1 Omega_(d)^((0))=1\Omega_{\mathrm{d}}^{(0)}=1Ωd(0)=1. Plugging in to eqn 17.33 we obtain
(E.145) t ( z ) = 1 H 0 0 ( z + 1 ) 1 d x x 1 2 = 2 3 H 0 ( z + 1 ) 3 2 (E.145) t ( z ) = 1 H 0 0 ( z + 1 ) 1 d x x 1 2 = 2 3 H 0 ( z + 1 ) 3 2 {:(E.145)t(z)=(1)/(H_(0))int_(0)^((z+1)^(-1))dxx^((1)/(2))=(2)/(3H_(0))(z+1)^(-(3)/(2)):}\begin{equation*} t(z)=\frac{1}{H_{0}} \int_{0}^{(z+1)^{-1}} \mathrm{~d} x x^{\frac{1}{2}}=\frac{2}{3 H_{0}}(z+1)^{-\frac{3}{2}} \tag{E.145} \end{equation*}(E.145)t(z)=1H00(z+1)1 dxx12=23H0(z+1)32
(18.1) (a) The first Friedmann equation is
(E.146) a ˙ 2 + k = 0 (E.147) a ˙ = 1 (E.146) a ˙ 2 + k = 0 (E.147) a ˙ = 1 {:[(E.146)a^(˙)^(2)+k=0],[(E.147)a^(˙)=1]:}\begin{gather*} \dot{a}^{2}+k=0 \tag{E.146}\\ \dot{a}=1 \tag{E.147} \end{gather*}(E.146)a˙2+k=0(E.147)a˙=1
so a ( t ) = t a ( t ) = t a(t)=ta(t)=ta(t)=t and the expression for the line element follows on setting k = 1 k = 1 k=-1k=-1k=1.
(b) We find
(E.148) d T 2 + d r 2 = d t 2 + t 2 d χ 2 (E.148) d T 2 + d r 2 = d t 2 + t 2 d χ 2 {:(E.148)-dT^(2)+dr^(2)=-dt^(2)+t^(2)dchi^(2):}\begin{equation*} -\mathrm{d} T^{2}+\mathrm{d} r^{2}=-\mathrm{d} t^{2}+t^{2} \mathrm{~d} \chi^{2} \tag{E.148} \end{equation*}(E.148)dT2+dr2=dt2+t2 dχ2
and so
d s 2 = d T 2 + d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = d T 2 + d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dT^(2)+dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)ds2=dT2+dr2+r2( dθ2+sin2θ dϕ2),
, (E.149)
which is the line element for empty Minkowski space (Universe 0).
(18.5) We have d X = T d T / X d X = T d T / X dX=TdT//X\mathrm{d} X=T \mathrm{~d} T / XdX=T dT/X, so
d s 2 = d T 2 + T 2 d T 2 1 + T 2 (E.150) = d T 2 1 + T 2 d s 2 = d T 2 + T 2 d T 2 1 + T 2 (E.150) = d T 2 1 + T 2 {:[ds^(2)=-dT^(2)+(T^(2)(d)T^(2))/(1+T^(2))],[(E.150)=-(dT^(2))/(1+T^(2))]:}\begin{align*} \mathrm{d} s^{2} & =-\mathrm{d} T^{2}+\frac{T^{2} \mathrm{~d} T^{2}}{1+T^{2}} \\ & =-\frac{\mathrm{d} T^{2}}{1+T^{2}} \tag{E.150} \end{align*}ds2=dT2+T2 dT21+T2(E.150)=dT21+T2
Writing T = sinh ψ T = sinh ψ T=sinh psiT=\sinh \psiT=sinhψ, which implies X = cosh ψ X = cosh ψ X=cosh psiX=\cosh \psiX=coshψ, we find d s 2 = d ψ 2 d s 2 = d ψ 2 ds^(2)=-dpsi^(2)\mathrm{d} s^{2}=-\mathrm{d} \psi^{2}ds2=dψ2.
Usually we would call ψ ψ psi\psiψ the proper time τ τ tau\tauτ. The embedding of a hyperbolic world line in Minkowski space therefore tells us that the intervals along the world line can be measured with the proper time, which is something we already knew.
(18.6) (a) Eliminate Y Y YYY to find
(E.151) d s 2 = d T 2 + d X 2 + ( T d T X d X ) 2 ( 1 + T 2 X 2 ) (E.151) d s 2 = d T 2 + d X 2 + ( T d T X d X ) 2 1 + T 2 X 2 {:(E.151)ds^(2)=-dT^(2)+dX^(2)+((T(d)T-X(d)X)^(2))/((1+T^(2)-X^(2))):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}+\frac{(T \mathrm{~d} T-X \mathrm{~d} X)^{2}}{\left(1+T^{2}-X^{2}\right)} \tag{E.151} \end{equation*}(E.151)ds2=dT2+dX2+(T dTX dX)2(1+T2X2)
Now define variables T = v cosh ψ T = v cosh ψ T=v cosh psiT=v \cosh \psiT=vcoshψ and X = v sinh ψ X = v sinh ψ X=v sinh psiX=v \sinh \psiX=vsinhψ to find
(E.152) d s 2 = d v 2 + v 2 d ψ 2 + v 2 d v 2 1 + v 2 (E.152) d s 2 = d v 2 + v 2 d ψ 2 + v 2 d v 2 1 + v 2 {:(E.152)ds^(2)=-dv^(2)+v^(2)dpsi^(2)+(v^(2)(d)v^(2))/(1+v^(2)):}\begin{equation*} \mathrm{d} s^{2}=-\mathrm{d} v^{2}+v^{2} \mathrm{~d} \psi^{2}+\frac{v^{2} \mathrm{~d} v^{2}}{1+v^{2}} \tag{E.152} \end{equation*}(E.152)ds2=dv2+v2 dψ2+v2 dv21+v2
which simplifies to the expression in the question.
(b) Set v = sinh χ v = sinh χ v=sinh chiv=\sinh \chiv=sinhχ to find that the relationship is
T = sinh χ cosh ψ , X = sinh χ sinh ψ , (E.153) Y = cosh χ T = sinh χ cosh ψ , X = sinh χ sinh ψ , (E.153) Y = cosh χ {:[T=sinh chi cosh psi","],[X=sinh chi sinh psi","],[(E.153)Y=cosh chi]:}\begin{align*} T & =\sinh \chi \cosh \psi, \\ X & =\sinh \chi \sinh \psi, \\ Y & =\cosh \chi \tag{E.153} \end{align*}T=sinhχcoshψ,X=sinhχsinhψ,(E.153)Y=coshχ
(c) The line element d s 2 = d χ 2 + cosh 2 χ d θ 2 d s 2 = d χ 2 + cosh 2 χ d θ 2 ds^(2)=-dchi^(2)+cosh^(2)chidtheta^(2)\mathrm{d} s^{2}=-\mathrm{d} \chi^{2}+\cosh ^{2} \chi \mathrm{~d} \theta^{2}ds2=dχ2+cosh2χ dθ2 is based on coordinates
T = sinh χ X = cosh χ cos θ (E.154) Y = cosh χ sin θ T = sinh χ X = cosh χ cos θ (E.154) Y = cosh χ sin θ {:[T=sinh chi],[X=cosh chi cos theta],[(E.154)Y=cosh chi sin theta]:}\begin{align*} T & =\sinh \chi \\ X & =\cosh \chi \cos \theta \\ Y & =\cosh \chi \sin \theta \tag{E.154} \end{align*}T=sinhχX=coshχcosθ(E.154)Y=coshχsinθ
This uses a set of circles in the X Y X Y X-YX-YXY plane as the basis for the coordinate system used. So if we fix the timelike variable χ χ chi\chiχ, the coordinates describe a circle in the X Y X Y X-YX-YXY plane. In contrast, a fixed time χ χ chi\chiχ in the other form of the line element gives us coordinates describing the hyperbolic curve T 2 + X 2 = 1 T 2 + X 2 = 1 -T^(2)+X^(2)=1-T^{2}+X^{2}=1T2+X2=1 in the X T X T X-TX-TXT plane. perbolic curve T 2 + X 2 = 1 T 2 + X 2 = 1 -T^(2)+X^(2)=1-T^{2}+X^{2}=1T2+X2=1 in the X T X T X-TX-TXT plane.
(d) In the absence of matter, we have a ˙ 2 + k = Λ a 2 / 3 a ˙ 2 + k = Λ a 2 / 3 a^(˙)^(2)+k=Lambdaa^(2)//3\dot{a}^{2}+k=\Lambda a^{2} / 3a˙2+k=Λa2/3. (d) In the absence of matter, we have a ˙ 2 + k = Λ a 2 / 3 a ˙ 2 + k = Λ a 2 / 3 a^(˙)^(2)+k=Lambdaa^(2)//3\dot{a}^{2}+k=\Lambda a^{2} / 3a˙2+k=Λa2/3.
In units where Λ / 3 = 1 Λ / 3 = 1 Lambda//3=1\Lambda / 3=1Λ/3=1, we have that a ( t ) = cosh t a ( t ) = cosh t a(t)=cosh ta(t)=\cosh ta(t)=cosht corresponds to a closed ( k = 1 ) ( k = 1 ) (k=1)(k=1)(k=1) universe, and a ( t ) = sinh t a ( t ) = sinh t a(t)=sinh ta(t)=\sinh ta(t)=sinht corresponds to an open universe.
(18.7) (b) Adding, we find T + X = e t T + X = e t T+X=e^(t)T+X=\mathrm{e}^{t}T+X=et or t = ln ( T + X ) t = ln ( T + X ) t=ln(T+X)t=\ln (T+X)t=ln(T+X). So the coordinates only cover the region T + X > 0 T + X > 0 T+X > 0T+X>0T+X>0. (c) Lines of constant t t ttt obey the equation T = A X T = A X T=A-XT=A-XT=AX, where A A AAA is a constant. They also obey T = B + C Y 2 T = B + C Y 2 T=B+CY^(2)T=B+C Y^{2}T=B+CY2, with B B BBB and C C CCC constants.
(d) The transformation needed here is u = e t u = e t u=e^(-t)u=\mathrm{e}^{-t}u=et.
(18.8) The embedding only works for x < 1 x < 1 x < 1x<1x<1. We have T + X = T + X = T+X=T+X=T+X= ( 1 x ) 1 2 e t > 0 ( 1 x ) 1 2 e t > 0 (1-x)^((1)/(2))e^(t) > 0(1-x)^{\frac{1}{2}} \mathrm{e}^{t}>0(1x)12et>0 and T + X = ( 1 x ) 1 2 e t > 0 T + X = ( 1 x ) 1 2 e t > 0 -T+X=(1-x)^((1)/(2))e^(-t) > 0-T+X=(1-x)^{\frac{1}{2}} \mathrm{e}^{-t}>0T+X=(1x)12et>0, so only the quarter of the hyperboloid with X | T | X | T | X >= |T|X \geq|T|X|T| is covered by these coordinates.
(19.2) (b) For β = 1 β = 1 beta=1\beta=1β=1 we have d u / d v = 0 d u / d v = 0 du//dv=0\mathrm{d} u / \mathrm{d} v=0du/dv=0, that we might call 'at rest' in terms of the coordinate velocity. For β = 0 β = 0 beta=0\beta=0β=0, the coordinate velocity d u / d v = 1 d u / d v = 1 du//dv=1\mathrm{d} u / \mathrm{d} v=1du/dv=1 ('light coordinate speed') and for β = 1 β = 1 beta=-1\beta=-1β=1 we have infinite coordinate speed. In short, we should beware of interpretations of coordinate velocities.
(19.3) (a) In component form, we have
(E.155) cos θ = g μ ν u μ u ν ( g α β u α u β g σ ρ v σ v ρ ) 1 2 . (E.155) cos θ = g μ ν u μ u ν g α β u α u β g σ ρ v σ v ρ 1 2 . {:(E.155)cos theta=(g_(mu nu)u^(mu)u^(nu))/((g_(alpha beta)u^(alpha)u^(beta)g_(sigma rho)v^(sigma)v^(rho))^((1)/(2))).:}\begin{equation*} \cos \theta=\frac{g_{\mu \nu} u^{\mu} u^{\nu}}{\left(g_{\alpha \beta} u^{\alpha} u^{\beta} g_{\sigma \rho} v^{\sigma} v^{\rho}\right)^{\frac{1}{2}}} . \tag{E.155} \end{equation*}(E.155)cosθ=gμνuμuν(gαβuαuβgσρvσvρ)12.
Substituting the transformation we find the factors of f ( x ) f ( x ) f(x)f(x)f(x) cancel, so the angle is unchanged.
(b) A null curve with tangent a a aaa has g ( a , a ) = 0 g ( a , a ) = 0 g(a,a)=0g(a, a)=0g(a,a)=0. In component form, we have
(E.156) 0 = g μ ν a μ a ν f ( x ) g μ ν a μ a ν (E.156) 0 = g μ ν a μ a ν f ( x ) g μ ν a μ a ν {:(E.156)0=g_(mu nu)a^(mu)a^(nu)rarr f(x)g_(mu nu)a^(mu)a^(nu):}\begin{equation*} 0=g_{\mu \nu} a^{\mu} a^{\nu} \rightarrow f(x) g_{\mu \nu} a^{\mu} a^{\nu} \tag{E.156} \end{equation*}(E.156)0=gμνaμaνf(x)gμνaμaν
which will still vanish, so the curve remains null.
(a) The form of the solution will be the same in the two spacetimes. The time dependence in terms of t t ttt is therefore written as
(E.157) A μ e i B d t a ( t ) (E.157) A μ e i B d t a ( t ) {:(E.157)A_(mu)prope^(iB int((d)t)/(a(t))):}\begin{equation*} A_{\mu} \propto \mathrm{e}^{\mathrm{i} B \int \frac{\mathrm{~d} t}{a(t)}} \tag{E.157} \end{equation*}(E.157)AμeiB dta(t)
(b) The instantaneous frequency ω ( t ) ω ( t ) omega(t)\omega(t)ω(t) can be extracted by taking its derivative of the phase of the wave with respect to time t t ttt, which tells us that ω ( t ) a ( t ) = ω ( t ) a ( t ) = omega(t)a(t)=\omega(t) a(t)=ω(t)a(t)= const.
(19.6) Putting the steps together we find
t = tan 1 ( t + r ) + tan 1 ( t r ) (E.158) r = tan 1 ( t + r ) tan 1 ( t r ) t = tan 1 ( t + r ) + tan 1 ( t r ) (E.158) r = tan 1 ( t + r ) tan 1 ( t r ) {:[t^(')=tan^(-1)(t+r)+tan^(-1)(t-r)],[(E.158)r^(')=tan^(-1)(t+r)-tan^(-1)(t-r)]:}\begin{align*} t^{\prime} & =\tan ^{-1}(t+r)+\tan ^{-1}(t-r) \\ r^{\prime} & =\tan ^{-1}(t+r)-\tan ^{-1}(t-r) \tag{E.158} \end{align*}t=tan1(t+r)+tan1(tr)(E.158)r=tan1(t+r)tan1(tr)
(19.7) The geodesics are shown in Fig. E.1.
Fig. E. 1 Penrose diagram for Exercise 19.7 showing (a) dotted, (b) solid and (c) dashed.
(19.8) (a) Use a ˙ 2 + k = C / a + Λ a 3 / 3 a ˙ 2 + k = C / a + Λ a 3 / 3 a^(˙)^(2)+k=C//a+Lambdaa^(3)//3\dot{a}^{2}+k=C / a+\Lambda a^{3} / 3a˙2+k=C/a+Λa3/3.
(b) Recall that a light ray travels a constant comoving distance
(E.159) t em t ob d t a ( t ) (E.159) t em t ob d t a ( t ) {:(E.159)int_(t_(em))^(t_(ob))((d)t)/(a(t)):}\begin{equation*} \int_{t_{\mathrm{em}}}^{t_{\mathrm{ob}}} \frac{\mathrm{~d} t}{a(t)} \tag{E.159} \end{equation*}(E.159)temtob dta(t)
and it is this distance that is measured by η η eta\etaη.
(c) Differentiating, we find
(E.160) d η d a = 2 C sin η = 1 a 1 2 ( C a ) 1 2 (E.160) d η d a = 2 C sin η = 1 a 1 2 ( C a ) 1 2 {:(E.160)(deta)/((d)a)=-(2)/(C sin eta)=(1)/(a^((1)/(2))(C-a)^((1)/(2))):}\begin{equation*} \frac{\mathrm{d} \eta}{\mathrm{~d} a}=-\frac{2}{C \sin \eta}=\frac{1}{a^{\frac{1}{2}}(C-a)^{\frac{1}{2}}} \tag{E.160} \end{equation*}(E.160)dη da=2Csinη=1a12(Ca)12
which is the equivalent to the equation of motion.
(d) To obtain the second equation, write
(E.161) d t = a ( t ) d η = C 2 ( 1 cos η ) d η (E.161) d t = a ( t ) d η = C 2 ( 1 cos η ) d η {:(E.161)dt=a(t)deta=(C)/(2)*(1-cos eta)deta:}\begin{equation*} \mathrm{d} t=a(t) \mathrm{d} \eta=\frac{C}{2} \cdot(1-\cos \eta) \mathrm{d} \eta \tag{E.161} \end{equation*}(E.161)dt=a(t)dη=C2(1cosη)dη
and integrate.
(e) The equations for a ( η ) a ( η ) a(eta)a(\eta)a(η) and t ( η ) t ( η ) t(eta)t(\eta)t(η) describe a cycloid for a ( t ) a ( t ) a(t)a(t)a(t). Starting at η = 0 η = 0 eta=0\eta=0η=0, the Universe reaches its maximum radius a = C a = C a=Ca=Ca=C when η = π η = π eta=pi\eta=\piη=π, and then reaches the Big Crunch at η = 2 π η = 2 π eta=2pi\eta=2 \piη=2π. The variable η η eta\etaη has a convenient interpretation in terms of the angle over which light traverses the unit circle, so at η = π η = π eta=pi\eta=\piη=π we have light travelling halfway around the circle from each direction. At the moment of the big crunch, when η = 2 π η = 2 π eta=2pi\eta=2 \piη=2π, light can therefore, only just, traverse the whole of the circle.
(20.3) (a) We have potential energy
U = d M ( r ) G m r U = d M ( r ) G m r U=-intdM(r)(Gm)/(r)U=-\int \mathrm{d} M(r) \frac{G m}{r}U=dM(r)Gmr
(E.162)
where d M ( r ) d M ( r ) dM(r)\mathrm{d} M(r)dM(r) is an element of mass separated from the particle by a distance r r rrr. For the ring we have all of the elements of mass at a distance r = ( a 2 + x 2 ) 1 2 r = a 2 + x 2 1 2 r=(a^(2)+x^(2))^((1)/(2))r=\left(a^{2}+x^{2}\right)^{\frac{1}{2}}r=(a2+x2)12 from m m mmm. We can at this point simply write the answer, or, if the mass per unit length is λ = M / 2 π a λ = M / 2 π a lambda=M//2pi a\lambda=M / 2 \pi aλ=M/2πa we can integrate elements of length a d θ a d θ adthetaa \mathrm{~d} \thetaa dθ to find
U = G M m 2 π a r 0 2 π a d θ = G M m r = G M m ( a 2 + x 1 2 ( E .163 ) . U = G M m 2 π a r 0 2 π a d θ = G M m r = G M m a 2 + x 1 2 ( E .163 ) . U=-(GMm)/(2pi ar)int_(0)^(2pi)adtheta=-(GMm)/(r)=-(GMm)/((a^(2)+x^((1)/(2))_((E.163)).)U=-\frac{G M m}{2 \pi a r} \int_{0}^{2 \pi} a \mathrm{~d} \theta=-\frac{G M m}{r}=-\frac{G M m}{\left(a^{2}+\underset{(\mathrm{E} .163)}{x^{\frac{1}{2}}} .\right.}U=GMm2πar02πa dθ=GMmr=GMm(a2+x12(E.163).
(b) By symmetry, the force can only act along the x x xxx direction (meaning F y = F z = 0 F y = F z = 0 F_(y)=F_(z)=0F_{y}=F_{z}=0Fy=Fz=0 ), so we differentiate with respect to x x xxx to find
(E.164) F x = G M m x ( a 2 + x 2 ) 3 2 . (E.164) F x = G M m x a 2 + x 2 3 2 . {:(E.164)F_(x)=-GMm(x)/((a^(2)+x^(2))^((3)/(2))).:}\begin{equation*} F_{x}=-G M m \frac{x}{\left(a^{2}+x^{2}\right)^{\frac{3}{2}}} . \tag{E.164} \end{equation*}(E.164)Fx=GMmx(a2+x2)32.
In the limit that x a x a x≫ax \gg axa, this becomes F x = G M m / x 2 F x = G M m / x 2 F_(x)=-GMm//x^(2)F_{x}=-G M m / x^{2}Fx=GMm/x2, the attractive force between two point masses.
(c) The disc may be built from concentric annular rings of radius a a aaa, width d a d a da\mathrm{d} ada and area 2 π a d a 2 π a d a 2pi ada2 \pi a \mathrm{~d} a2πa da. If we take the total mass per unit area of the disc to be σ = Ω / ( π L 2 ) σ = Ω / π L 2 sigma=Omega//(piL^(2))\sigma=\Omega /\left(\pi L^{2}\right)σ=Ω/(πL2), then each annular element has mass d Ω = 2 π a d a σ d Ω = 2 π a d a σ dOmega=2pi ada sigma\mathrm{d} \Omega=2 \pi a \mathrm{~d} a \sigmadΩ=2πa daσ, and the contribution to the potential energy from each element is
(E.165) d U = G m d Ω ( a 2 + x 2 ) 1 2 = Gm 2 π σ a d a ( a 2 + x 2 ) 1 2 . (E.165) d U = G m d Ω a 2 + x 2 1 2 = Gm 2 π σ a d a a 2 + x 2 1 2 . {:(E.165)dU=-(Gm(d)Omega)/((a^(2)+x^(2))^((1)/(2)))=-Gm2pi sigma(a(d)a)/((a^(2)+x^(2))^((1)/(2))).:}\begin{equation*} \mathrm{d} U=-\frac{G m \mathrm{~d} \Omega}{\left(a^{2}+x^{2}\right)^{\frac{1}{2}}}=-\operatorname{Gm} 2 \pi \sigma \frac{a \mathrm{~d} a}{\left(a^{2}+x^{2}\right)^{\frac{1}{2}}} . \tag{E.165} \end{equation*}(E.165)dU=Gm dΩ(a2+x2)12=Gm2πσa da(a2+x2)12.
Integrating and substituting for σ σ sigma\sigmaσ yields
(E.166) U = 2 G m Ω L 2 [ ( L 2 + x 2 ) 1 2 x ] (E.166) U = 2 G m Ω L 2 L 2 + x 2 1 2 x {:(E.166)U=-(2Gm Omega)/(L^(2))[(L^(2)+x^(2))^((1)/(2))-x]:}\begin{equation*} U=-\frac{2 G m \Omega}{L^{2}}\left[\left(L^{2}+x^{2}\right)^{\frac{1}{2}}-x\right] \tag{E.166} \end{equation*}(E.166)U=2GmΩL2[(L2+x2)12x]
(20.4) (a) Multiply through by a velocity component u α u α u^(alpha)u^{\alpha}uα and sum to find
(E.167) m u α a α = m Φ x α u α (E.167) m u α a α = m Φ x α u α {:(E.167)mu^(alpha)a_(alpha)=-m(del Phi)/(delx^(alpha))*u^(alpha):}\begin{equation*} m u^{\alpha} a_{\alpha}=-m \frac{\partial \Phi}{\partial x^{\alpha}} \cdot u^{\alpha} \tag{E.167} \end{equation*}(E.167)muαaα=mΦxαuα
The right-hand side will not vanish in general.
(b) Multiplying through by u α u α u_(alpha)u_{\alpha}uα, the bracketed part becomes
(E.168) u α ( η α + u α u α u β ) (E.168) u α η α + u α u α u β {:(E.168)u_(alpha)(eta^(alpha)+u_(alpha)u^(alpha)u^(beta)):}\begin{equation*} u_{\alpha}\left(\eta^{\alpha}+u_{\alpha} u^{\alpha} u^{\beta}\right) \tag{E.168} \end{equation*}(E.168)uα(ηα+uαuαuβ)
which, noting that u α u α = 1 u α u α = 1 u_(alpha)u^(alpha)=-1u_{\alpha} u^{\alpha}=-1uαuα=1, gives u β u β = 0 u β u β = 0 u^(beta)-u^(beta)=0u^{\beta}-u^{\beta}=0uβuβ=0.
(20.5) The equation of motion can be rewritten as
(E.169) d p α d x β d x β d λ 1 m = m ( η α β + u α u β ) Φ x β (E.169) d p α d x β d x β d λ 1 m = m η α β + u α u β Φ x β {:(E.169)(dp^(alpha))/(dx^(beta))*(dx^(beta))/(dlambda)*(1)/(m)=-m(eta^(alpha beta)+u^(alpha)u^(beta))(del Phi)/(delx^(beta)):}\begin{equation*} \frac{\mathrm{d} p^{\alpha}}{\mathrm{d} x^{\beta}} \cdot \frac{\mathrm{d} x^{\beta}}{\mathrm{d} \lambda} \cdot \frac{1}{m}=-m\left(\eta^{\alpha \beta}+u^{\alpha} u^{\beta}\right) \frac{\partial \Phi}{\partial x^{\beta}} \tag{E.169} \end{equation*}(E.169)dpαdxβdxβdλ1m=m(ηαβ+uαuβ)Φxβ
( d p α d x β + p α Φ x β ) d x β d λ 1 m = m η α β Φ x β d p α d x β + p α Φ x β d x β d λ 1 m = m η α β Φ x β ((dp^(alpha))/(dx^(beta))+p^(alpha)(del Phi)/(delx^(beta)))(dx^(beta))/(dlambda)*(1)/(m)=-meta^(alpha beta)(del Phi)/(delx^(beta))\left(\frac{\mathrm{d} p^{\alpha}}{\mathrm{d} x^{\beta}}+p^{\alpha} \frac{\partial \Phi}{\partial x^{\beta}}\right) \frac{\mathrm{d} x^{\beta}}{\mathrm{d} \lambda} \cdot \frac{1}{m}=-m \eta^{\alpha \beta} \frac{\partial \Phi}{\partial x^{\beta}}(dpαdxβ+pαΦxβ)dxβdλ1m=mηαβΦxβ.
The right-hand side vanishes as m 0 m 0 m rarr0m \rightarrow 0m0 leaving
(E.170) d p α d x β + p α Φ x β = 0 (E.170) d p α d x β + p α Φ x β = 0 {:(E.170)(dp^(alpha))/(dx^(beta))+p^(alpha)(del Phi)/(delx^(beta))=0:}\begin{equation*} \frac{\mathrm{d} p^{\alpha}}{\mathrm{d} x^{\beta}}+p^{\alpha} \frac{\partial \Phi}{\partial x^{\beta}}=0 \tag{E.170} \end{equation*}(E.170)dpαdxβ+pαΦxβ=0
whose solution is p α e Φ = p α e Φ = p^(alpha)e^(Phi)=p^{\alpha} e^{\Phi}=pαeΦ= (const) for all of the coordinates. As a result, the ratios of the components of the momentum remain constant along the world line. There can therefore be no deflection of the light rays by the field.
(21.1) We find
(E.171) T t t t t = ρ e 2 Φ , T r r = p e 2 Λ T θ θ = p r 2 , T ϕ ϕ = p r 2 sin θ (E.171) T t t t t = ρ e 2 Φ , T r r = p e 2 Λ T θ θ = p r 2 , T ϕ ϕ = p r 2 sin θ {:(E.171){:[T_(tt)^(tt)=rhoe^(-2Phi)",",T^(rr)=pe^(-2Lambda)],[T^(theta theta)=(p)/(r^(2))",",T^(phi phi)=(p)/(r^(2)sin theta)]:}:}\begin{array}{cc} T_{t t}^{t t}=\rho \mathrm{e}^{-2 \Phi}, & T^{r r}=p \mathrm{e}^{-2 \Lambda} \tag{E.171}\\ T^{\theta \theta}=\frac{p}{r^{2}}, & T^{\phi \phi}=\frac{p}{r^{2} \sin \theta} \end{array}(E.171)Ttttt=ρe2Φ,Trr=pe2ΛTθθ=pr2,Tϕϕ=pr2sinθ
(21.5) (a) We find g = r 2 sin θ g = r 2 sin θ g=r^(2)sin thetag=r^{2} \sin \thetag=r2sinθ and so the area is 4 π r 2 4 π r 2 4pir^(2)4 \pi r^{2}4πr2. (b) The circumference is 2 π r sin θ 2 π r sin θ 2pi r sin theta2 \pi r \sin \theta2πrsinθ.
(21.6) (a) Consider a spherical mass of fluid and a sphericalshell mass element of the fluid at a radius r r rrr, with surface shell mass element of the fluid at a radius r r rrr, with surface
area A A AAA, mass d m d m dm\mathrm{d} mdm and width d r d r dr\mathrm{d} rdr. There is an outwardarea A A AAA, mass d m d m dm\mathrm{d} mdm and width d r d r dr\mathrm{d} rdr. There is an outward-
directed force on the inner surface of the element of p ( r ) A p ( r ) A p(r)Ap(r) Ap(r)A, resulting from the fluid closer to the centre. There are inward forces from (i) fluid outside the element, pushing back on the outer surface with force p ( r + d r ) A p ( r + d r ) A p(r+dr)Ap(r+\mathrm{d} r) Ap(r+dr)A; and (ii) the gravitational pull of the fluid closer to the centre of Gm ( r ) d m / r 2 Gm ( r ) d m / r 2 Gm(r)dm//r^(2)\operatorname{Gm}(r) \mathrm{d} m / r^{2}Gm(r)dm/r2 (i.e. the fluid at a smaller radius acts as if it is all concentrated at the
origin in Newtonian gravitation). Equating forces we have
(E.172) A p ( r ) = A p ( r + d r ) + G m ( r ) d m r 2 (E.172) A p ( r ) = A p ( r + d r ) + G m ( r ) d m r 2 {:(E.172)Ap(r)=Ap(r+dr)+(Gm(r)dm)/(r^(2)):}\begin{equation*} A p(r)=A p(r+\mathrm{d} r)+\frac{G m(r) \mathrm{d} m}{r^{2}} \tag{E.172} \end{equation*}(E.172)Ap(r)=Ap(r+dr)+Gm(r)dmr2
Rearranging gives
(E.173) A d p = G m ( r ) d m r 2 (E.173) A d p = G m ( r ) d m r 2 {:(E.173)Adp=-(Gm(r)dm)/(r^(2)):}\begin{equation*} A \mathrm{~d} p=-\frac{G m(r) \mathrm{d} m}{r^{2}} \tag{E.173} \end{equation*}(E.173)A dp=Gm(r)dmr2
where d p = p ( r + d r ) p ( r ) d p = p ( r + d r ) p ( r ) dp=p(r+dr)-p(r)\mathrm{d} p=p(r+\mathrm{d} r)-p(r)dp=p(r+dr)p(r). Finally we note that A = 4 π r 2 A = 4 π r 2 A=4pir^(2)A=4 \pi r^{2}A=4πr2 and d m = ρ A d r d m = ρ A d r dm=rho Adr\mathrm{d} m=\rho A \mathrm{~d} rdm=ρA dr, where ρ ρ rho\rhoρ is the density and we write m ( r ) = M m ( r ) = M m(r)=Mm(r)=Mm(r)=M.
(21.7) (b) Set G r ^ t ^ = 0 G r ^ t ^ = 0 G_( hat(r) hat(t))=0G_{\hat{r} \hat{t}}=0Gr^t^=0 to find Λ , t = 0 Λ , t = 0 Lambda_(,t)=0\Lambda_{, t}=0Λ,t=0.
(c) Set G t ^ t ^ = 0 G t ^ t ^ = 0 G_( hat(t) hat(t))=0G_{\hat{t} \hat{t}}=0Gt^t^=0 to find
(E.174) d r r = 2 d Λ 1 e 2 Λ (E.174) d r r = 2 d Λ 1 e 2 Λ {:(E.174)(dr)/(r)=(2(d)Lambda)/(1-e^(2Lambda)):}\begin{equation*} \frac{\mathrm{d} r}{r}=\frac{2 \mathrm{~d} \Lambda}{1-\mathrm{e}^{2 \Lambda}} \tag{E.174} \end{equation*}(E.174)drr=2 dΛ1e2Λ
Integrate this to find
(E.175) r = C e 2 Λ 1 , (E.175) r = C e 2 Λ 1 , {:(E.175)r=(C)/(e^(-2Lambda)-1)",":}\begin{equation*} r=\frac{C}{\mathrm{e}^{-2 \Lambda}-1}, \tag{E.175} \end{equation*}(E.175)r=Ce2Λ1,
and choose C = 2 M C = 2 M C=-2MC=-2 MC=2M.
(d) Consider G r ^ r ^ G i t ^ = 0 G r ^ r ^ G i t ^ = 0 G_( hat(r) hat(r))-G_(i hat(t))=0G_{\hat{r} \hat{r}}-G_{i \hat{t}}=0Gr^r^Git^=0.
(e) Rescale d t = e f ( t ) d t d t = e f ( t ) d t dt^(')=e^(f(t))dt\mathrm{d} t^{\prime}=\mathrm{e}^{f(t)} \mathrm{d} tdt=ef(t)dt and the Schwarzschild metric is recovered.
(21.8) (a) Compare eqns 21.12 and 21.17, writing ρ = p = ζ ρ = p = ζ rho=-p=zeta\rho=-p=\zetaρ=p=ζ. (b) Substitution reveals we have a solution if H 2 = H 2 = H^(2)=H^{2}=H2= 8 π ζ / 3 8 π ζ / 3 8pi zeta//38 \pi \zeta / 38πζ/3, which is the result for the Hubble constant in the de Sitter Universe.
(22.1) (a) The velocity has components
(E.176) [ ( g 00 ) 1 2 , 0 , 0 , 0 ] (E.176) g 00 1 2 , 0 , 0 , 0 {:(E.176)[(-g_(00))^(-(1)/(2)),0,0,0]:}\begin{equation*} \left[\left(-g_{00}\right)^{-\frac{1}{2}}, 0,0,0\right] \tag{E.176} \end{equation*}(E.176)[(g00)12,0,0,0]
with g 00 g 00 g_(00)g_{00}g00 static, so the acceleration becomes
(E.177) a μ = ( u 0 ) 2 Γ 00 μ (E.177) a μ = u 0 2 Γ 00 μ {:(E.177)a^(mu)=(u^(0))^(2)Gamma_(00)^(mu):}\begin{equation*} a^{\mu}=\left(u^{0}\right)^{2} \Gamma_{00}^{\mu} \tag{E.177} \end{equation*}(E.177)aμ=(u0)2Γ00μ
where the connection is given by Γ μ 00 = 1 2 g 00 , σ g μ σ Γ μ 00 = 1 2 g 00 , σ g μ σ Gamma^(mu)_(00)=-(1)/(2)g_(00,sigma)g^(mu sigma)\Gamma^{\mu}{ }_{00}=-\frac{1}{2} g_{00, \sigma} g^{\mu \sigma}Γμ00=12g00,σgμσ. We then have a 0 = 0 a 0 = 0 a^(0)=0a^{0}=0a0=0 and
(E.178) a i = 1 2 ( g i j g 00 ) g 00 , j (E.178) a i = 1 2 g i j g 00 g 00 , j {:(E.178)a^(i)=(1)/(2)((g^(ij))/(g_(00)))g_(00,j):}\begin{equation*} a^{i}=\frac{1}{2}\left(\frac{g^{i j}}{g_{00}}\right) g_{00, j} \tag{E.178} \end{equation*}(E.178)ai=12(gijg00)g00,j
(b) Plugging in the components of the Schwarzschild metric we obtain g 00 , 1 = 2 M / r 2 g 00 , 1 = 2 M / r 2 g_(00,1)=-2M//r^(2)g_{00,1}=-2 M / r^{2}g00,1=2M/r2 and so a μ = a μ = a^(mu)=a^{\mu}=aμ= ( 0 , M / r 2 , 0 , 0 ) 0 , M / r 2 , 0 , 0 (0,M//r^(2),0,0)\left(0, M / r^{2}, 0,0\right)(0,M/r2,0,0). That is, despite not moving, the particle is accelerating outwards as it is not following a ticle is ac
geodesic.
geodesic.
(c) Recall from Chapter 2 that the square of the proper acceleration α 2 α 2 alpha^(2)\alpha^{2}α2 is equal to a 2 = ( 1 2 M / r ) 1 ( M 2 / r 4 ) a 2 = ( 1 2 M / r ) 1 M 2 / r 4 a^(2)=(1-2M//r)^(-1)(M^(2)//r^(4))\boldsymbol{a}^{2}=(1-2 M / r)^{-1}\left(M^{2} / r^{4}\right)a2=(12M/r)1(M2/r4), so α α alpha\alphaα is given by
(E.179) α = M r 2 ( r r 2 M ) 1 2 (E.179) α = M r 2 r r 2 M 1 2 {:(E.179)alpha=(M)/(r^(2))((r)/(r-2M))^((1)/(2)):}\begin{equation*} \alpha=\frac{M}{r^{2}}\left(\frac{r}{r-2 M}\right)^{\frac{1}{2}} \tag{E.179} \end{equation*}(E.179)α=Mr2(rr2M)12
(22.2) (a) The equations of motion are as follows: (i) For the t t ttt coordinate
(E.180) t ¨ + 2 M r ( r 2 M ) r ˙ t ˙ = 0 (E.180) t ¨ + 2 M r ( r 2 M ) r ˙ t ˙ = 0 {:(E.180)t^(¨)+(2M)/(r(r-2M))r^(˙)t^(˙)=0:}\begin{equation*} \ddot{t}+\frac{2 M}{r(r-2 M)} \dot{r} \dot{t}=0 \tag{E.180} \end{equation*}(E.180)t¨+2Mr(r2M)r˙t˙=0
(ii) For the r r rrr coordinate
r ¨ M r ( r 2 M ) r ˙ 2 + M ( r 2 M ) r 3 t ˙ 2 ( r 2 M ) θ ˙ 2 (E.181) ( r 2 M ) sin 2 θ ϕ ˙ 2 = 0 . r ¨ M r ( r 2 M ) r ˙ 2 + M ( r 2 M ) r 3 t ˙ 2 ( r 2 M ) θ ˙ 2 (E.181) ( r 2 M ) sin 2 θ ϕ ˙ 2 = 0 . {:[r^(¨)-(M)/(r(r-2M))r^(˙)^(2)+(M(r-2M))/(r^(3))t^(˙)^(2)-(r-2M)theta^(˙)^(2)],[(E.181)-(r-2M)sin^(2)thetaphi^(˙)^(2)=0.]:}\begin{align*} & \ddot{r}-\frac{M}{r(r-2 M)} \dot{r}^{2}+\frac{M(r-2 M)}{r^{3}} \dot{t}^{2}-(r-2 M) \dot{\theta}^{2} \\ & -(r-2 M) \sin ^{2} \theta \dot{\phi}^{2}=0 . \tag{E.181} \end{align*}r¨Mr(r2M)r˙2+M(r2M)r3t˙2(r2M)θ˙2(E.181)(r2M)sin2θϕ˙2=0.
(iii) For the θ θ theta\thetaθ coordinate
(E.182) θ ¨ + 2 r r ˙ θ ˙ sin θ cos θ ϕ ˙ 2 = 0 (E.182) θ ¨ + 2 r r ˙ θ ˙ sin θ cos θ ϕ ˙ 2 = 0 {:(E.182)theta^(¨)+(2)/(r)r^(˙)theta^(˙)-sin theta cos thetaphi^(˙)^(2)=0:}\begin{equation*} \ddot{\theta}+\frac{2}{r} \dot{r} \dot{\theta}-\sin \theta \cos \theta \dot{\phi}^{2}=0 \tag{E.182} \end{equation*}(E.182)θ¨+2rr˙θ˙sinθcosθϕ˙2=0
(iv) For the ϕ ϕ phi\phiϕ coordinate
(E.183) ϕ ¨ + 2 r r ˙ ϕ ˙ + 2 cos θ sin θ θ ˙ ϕ ˙ = 0 (E.183) ϕ ¨ + 2 r r ˙ ϕ ˙ + 2 cos θ sin θ θ ˙ ϕ ˙ = 0 {:(E.183)phi^(¨)+(2)/(r)r^(˙)phi^(˙)+(2cos theta)/(sin theta)theta^(˙)phi^(˙)=0:}\begin{equation*} \ddot{\phi}+\frac{2}{r} \dot{r} \dot{\phi}+\frac{2 \cos \theta}{\sin \theta} \dot{\theta} \dot{\phi}=0 \tag{E.183} \end{equation*}(E.183)ϕ¨+2rr˙ϕ˙+2cosθsinθθ˙ϕ˙=0
(b) Set θ = π / 2 θ = π / 2 theta=pi//2\theta=\pi / 2θ=π/2 and θ ˙ = 0 θ ˙ = 0 theta^(˙)=0\dot{\theta}=0θ˙=0. We obtain the following, in terms of the conserved quantities:
(i) For the t t ttt coordinate
(E.184) d d τ E ~ = 0 (E.184) d d τ E ~ = 0 {:(E.184)(d)/((d)tau)* tilde(E)=0:}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau} \cdot \tilde{E}=0 \tag{E.184} \end{equation*}(E.184)d dτE~=0
(ii) For the r r rrr coordinate
r ¨ M r ( r 2 M ) r ˙ 2 + M E ~ 2 r ( r 2 M ) ( r 2 M ) L ~ 2 r 4 = 0 r ¨ M r ( r 2 M ) r ˙ 2 + M E ~ 2 r ( r 2 M ) ( r 2 M ) L ~ 2 r 4 = 0 r^(¨)-(M)/(r(r-2M))r^(˙)^(2)+(M tilde(E)^(2))/(r(r-2M))-((r-2M) tilde(L)^(2))/(r^(4))=0\ddot{r}-\frac{M}{r(r-2 M)} \dot{r}^{2}+\frac{M \tilde{E}^{2}}{r(r-2 M)}-\frac{(r-2 M) \tilde{L}^{2}}{r^{4}}=0r¨Mr(r2M)r˙2+ME~2r(r2M)(r2M)L~2r4=0
(iii) For the θ θ theta\thetaθ coordinate θ ¨ = 0 θ ¨ = 0 theta^(¨)=0\ddot{\theta}=0θ¨=0.
(iv) For the ϕ ϕ phi\phiϕ coordinate
(E.186) d d τ L ~ = 0 (E.186) d d τ L ~ = 0 {:(E.186)(d)/((d)tau) tilde(L)=0:}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau} \tilde{L}=0 \tag{E.186} \end{equation*}(E.186)d dτL~=0
(c) Substituting r ˙ r ˙ r^(˙)\dot{r}r˙ from the effective-energy equation into the equation of motion for r ¨ r ¨ r^(¨)\ddot{r}r¨ gives a simplified equation of motion
(E.187) r ¨ + M r 2 + 3 M L ~ 2 r 4 L ~ 2 r 3 = 0 (E.187) r ¨ + M r 2 + 3 M L ~ 2 r 4 L ~ 2 r 3 = 0 {:(E.187)r^(¨)+(M)/(r^(2))+(3M tilde(L)^(2))/(r^(4))-( tilde(L)^(2))/(r^(3))=0:}\begin{equation*} \ddot{r}+\frac{M}{r^{2}}+\frac{3 M \tilde{L}^{2}}{r^{4}}-\frac{\tilde{L}^{2}}{r^{3}}=0 \tag{E.187} \end{equation*}(E.187)r¨+Mr2+3ML~2r4L~2r3=0
Differentiating the effective energy equation with respect to proper time gives an identical equation, so these are consistent.
(22.3) Using the conservation of p t p t p_(t)p_{t}pt along the geodesic, we can relate the values of p t p t p^(t)p^{t}pt at radius R R RRR and at oo\infty by raising the index and writing g t t g t t g_(tt)g_{t t}gtt on both sides
(E.188) ( 1 2 M R ) p t ( R ) = p t ( ) (E.188) 1 2 M R p t ( R ) = p t ( ) {:(E.188)(1-(2M)/(R))p^(t)(R)=p^(t)(oo):}\begin{equation*} \left(1-\frac{2 M}{R}\right) p^{t}(R)=p^{t}(\infty) \tag{E.188} \end{equation*}(E.188)(12MR)pt(R)=pt()
We are looking to write this in terms of measured values p t ^ p t ^ p^( hat(t))p^{\hat{t}}pt^ [where p t = ( e t ^ ) t p t ] p t = e t ^ t p t {:p^(t)=(e_( hat(t)))^(t)p^(t)]\left.p^{t}=\left(e_{\hat{t}}\right)^{t} p^{t}\right]pt=(et^)tpt], so we use the vielbein component ( e t ^ ) t = ( 1 2 M / r ) 1 2 e t ^ t = ( 1 2 M / r ) 1 2 (e_( hat(t)))^(t)=(1-2M//r)^(-(1)/(2))\left(e_{\hat{t}}\right)^{t}=(1-2 M / r)^{-\frac{1}{2}}(et^)t=(12M/r)12 so we have
(E.189) ( 1 2 M R ) 1 2 p t ^ ( R ) = p t ^ ( ) (E.189) 1 2 M R 1 2 p t ^ ( R ) = p t ^ ( ) {:(E.189)(1-(2M)/(R))^((1)/(2))p^( hat(t))(R)=p^( hat(t))(oo):}\begin{equation*} \left(1-\frac{2 M}{R}\right)^{\frac{1}{2}} p^{\hat{t}}(R)=p^{\hat{t}}(\infty) \tag{E.189} \end{equation*}(E.189)(12MR)12pt^(R)=pt^()
Setting p t ^ ( R ) = ω R p t ^ ( R ) = ω R p^( hat(t))(R)=ℏomega_(R)p^{\hat{t}}(R)=\hbar \omega_{R}pt^(R)=ωR and p t ^ ( ) = ω p t ^ ( ) = ω p^( hat(t))(oo)=ℏomega_(oo)p^{\hat{t}}(\infty)=\hbar \omega_{\infty}pt^()=ω, the answer follows.
(22.5) (a) The escape velocity is defined such that the velocity u r u r u^(r)u^{r}ur at r = r = r=oor=\inftyr= is zero. At r = r = r=oor=\inftyr= we have E ~ = 1 E ~ = 1 tilde(E)=1\tilde{E}=1E~=1, which is then true along the geodesic. This E = 1 E = 1 E=1E=1E=1, which is then true along the geodesic. This
gives u t = ( 1 2 M / r ) 1 u t = ( 1 2 M / r ) 1 u^(t)=(1-2M//r)^(-1)u^{t}=(1-2 M / r)^{-1}ut=(12M/r)1. The effective energy equation, or u μ u μ = 1 u μ u μ = 1 u^(mu)u_(mu)=-1u^{\mu} u_{\mu}=-1uμuμ=1, then gives u r = ( 2 M / r ) 1 2 u r = ( 2 M / r ) 1 2 u^(r)=(2M//r)^((1)/(2))u^{r}=(2 M / r)^{\frac{1}{2}}ur=(2M/r)12. The coordinate velocity at r = R r = R r=Rr=Rr=R is then
d r d t = u r u t = ( 2 M R ) 1 2 ( 1 2 M R ) d r d t = u r u t = 2 M R 1 2 1 2 M R (dr)/((d)t)=(u^(r))/(u^(t))=((2M)/(R))^((1)/(2))(1-(2M)/(R))\frac{\mathrm{d} r}{\mathrm{~d} t}=\frac{u^{r}}{u^{t}}=\left(\frac{2 M}{R}\right)^{\frac{1}{2}}\left(1-\frac{2 M}{R}\right)dr dt=urut=(2MR)12(12MR)
The observer makes observations in their rest frame and so, using the vielbein components in the chapter, we have
d r ^ d t ^ = ( e r ) r ^ ( e t ) t ^ d r d t = ( 1 2 M R ) 1 d r d t = ( 2 M R ) (E. 191 1 2 d r ^ d t ^ = e r r ^ e t t ^ d r d t = 1 2 M R 1 d r d t = 2 M R (E.  191 1 2 (d( hat(r)))/((d)( hat(t)))=((e_(r))^( hat(r)))/((e_(t))^( hat(t)))((d)r)/((d)t)=(1-(2M)/(R))^(-1)((d)r)/((d)t)=((2M)/(R))_((E. 191)^((1)/(2))\frac{\mathrm{d} \hat{r}}{\mathrm{~d} \hat{t}}=\frac{\left(\boldsymbol{e}_{r}\right)^{\hat{r}}}{\left(\boldsymbol{e}_{t}\right)^{\hat{t}}} \frac{\mathrm{~d} r}{\mathrm{~d} t}=\left(1-\frac{2 M}{R}\right)^{-1} \frac{\mathrm{~d} r}{\mathrm{~d} t}=\left(\frac{2 M}{R}\right)_{\text {(E. } 191}^{\frac{1}{2}}dr^ dt^=(er)r^(et)t^ dr dt=(12MR)1 dr dt=(2MR)(E. 19112
This is the same as the non-relativistic result.
(b) The energy measured by an observer is E = m u E = m u E=-muE=-m \boldsymbol{u}E=mu u obs u obs  u_("obs ")\boldsymbol{u}_{\text {obs }}uobs  or (more simply) p t ^ = m u t ^ = m u t ( e t ) t ^ p t ^ = m u t ^ = m u t e t t ^ p^( hat(t))=mu^( hat(t))=mu^(t)(e_(t))^( hat(t))p^{\hat{t}}=m u^{\hat{t}}=m u^{t}\left(\boldsymbol{e}_{t}\right)^{\hat{t}}pt^=mut^=mut(et)t^. These give
(E.192) p t ^ = m ( 1 2 M R ) 1 2 (E.192) p t ^ = m 1 2 M R 1 2 {:(E.192)p^( hat(t))=m(1-(2M)/(R))^(-(1)/(2)):}\begin{equation*} p^{\hat{t}}=m\left(1-\frac{2 M}{R}\right)^{-\frac{1}{2}} \tag{E.192} \end{equation*}(E.192)pt^=m(12MR)12
(22.6) (a) Write the angular speed of the earth as ω = d ϕ / d t ω = d ϕ / d t omega=dphi//dt\omega=\mathrm{d} \phi / \mathrm{d} tω=dϕ/dt and we obtain, restoring factors,
(E.193) d τ 1 = ( 1 2 G M c 2 r r 2 ω 2 c 2 ) 1 2 d t (E.193) d τ 1 = 1 2 G M c 2 r r 2 ω 2 c 2 1 2 d t {:(E.193)dtau_(1)=(1-(2GM)/(c^(2)r)-(r^(2)omega^(2))/(c^(2)))^((1)/(2))dt:}\begin{equation*} \mathrm{d} \tau_{1}=\left(1-\frac{2 G M}{c^{2} r}-\frac{r^{2} \omega^{2}}{c^{2}}\right)^{\frac{1}{2}} \mathrm{~d} t \tag{E.193} \end{equation*}(E.193)dτ1=(12GMc2rr2ω2c2)12 dt
The second and third terms are small and so the answer follows on expanding the bracket to first order.
(b) Make the substitution r r + h r r + h r rarr r+hr \rightarrow r+hrr+h and ω r ω r omega r rarr\omega r \rightarrowωr ω ( r + h ) + v ω ( r + h ) + v omega(r+h)+v\omega(r+h)+vω(r+h)+v and expand, noting that h r h r h≪rh \ll rhr.
(c) Restoring factors we have that
(E.194) Δ G M h c 2 r 2 + v 2 c 2 ( v + 2 ω r ) (E.194) Δ G M h c 2 r 2 + v 2 c 2 ( v + 2 ω r ) {:(E.194)Delta~~-(GMh)/(c^(2)r^(2))+(v)/(2c^(2))(v+2omega r):}\begin{equation*} \Delta \approx-\frac{G M h}{c^{2} r^{2}}+\frac{v}{2 c^{2}}(v+2 \omega r) \tag{E.194} \end{equation*}(E.194)ΔGMhc2r2+v2c2(v+2ωr)
Take v = 250 ms 1 , h = 10 4 m v = 250 ms 1 , h = 10 4 m v=250ms^(-1),h=10^(4)mv=250 \mathrm{~ms}^{-1}, h=10^{4} \mathrm{~m}v=250 ms1,h=104 m and ω = 10 4 rads 1 ω = 10 4 rads 1 omega=10^(-4)rads^(-1)\omega=10^{-4} \mathrm{rads}^{-1}ω=104rads1, and note that G M / r 2 10 ms 2 G M / r 2 10 ms 2 GM//r^(2)~~10ms^(-2)G M / r^{2} \approx 10 \mathrm{~ms}^{-2}GM/r210 ms2, to find Δ 1 × 10 12 Δ 1 × 10 12 Delta~~1xx10^(-12)\Delta \approx 1 \times 10^{-12}Δ1×1012 for eastward flight.
(d) For westward flight substitute v v v v v rarr-vv \rightarrow-vvv to find Δ 2 × 10 12 Δ 2 × 10 12 Delta~~-2xx10^(-12)\Delta \approx-2 \times 10^{-12}Δ2×1012.
(22.7) The observer at rest has velocity u u u\boldsymbol{u}u with components u μ = ( u t , 0 , 0 , 0 ) u μ = u t , 0 , 0 , 0 u^(mu)=(u^(t),0,0,0)u^{\mu}=\left(u^{t}, 0,0,0\right)uμ=(ut,0,0,0), from which we can use the constraint on velocity to evaluate
(E.195) g t t ( r 1 ) ( u t ) 2 = 1 (E.195) g t t r 1 u t 2 = 1 {:(E.195)g_(tt)(r_(1))(u^(t))^(2)=-1:}\begin{equation*} g_{t t}\left(r_{1}\right)\left(u^{t}\right)^{2}=-1 \tag{E.195} \end{equation*}(E.195)gtt(r1)(ut)2=1
giving
(E.196) u t = ( 1 2 M r 1 ) 1 2 (E.196) u t = 1 2 M r 1 1 2 {:(E.196)u^(t)=(1-(2M)/(r_(1)))^(-(1)/(2)):}\begin{equation*} u^{t}=\left(1-\frac{2 M}{r_{1}}\right)^{-\frac{1}{2}} \tag{E.196} \end{equation*}(E.196)ut=(12Mr1)12
Since we are planning to take a dot product, we only need to evaluate the timelike component of the other observer's velocity v v v\boldsymbol{v}v. This observer is free falling, and so we have that E ~ = v t ( r 2 ) = v t ( r 1 ) E ~ = v t r 2 = v t r 1 - tilde(E)=v_(t)(r_(2))=v_(t)(r_(1))-\tilde{E}=v_{t}\left(r_{2}\right)=v_{t}\left(r_{1}\right)E~=vt(r2)=vt(r1). Since the observer
so
(E.197) v t ( r 2 ) = ( 1 2 M r 2 ) 1 2 = v t ( r 1 ) (E.197) v t r 2 = 1 2 M r 2 1 2 = v t r 1 {:(E.197)v_(t)(r_(2))=-(1-(2M)/(r_(2)))^((1)/(2))=v_(t)(r_(1)):}\begin{equation*} v_{t}\left(r_{2}\right)=-\left(1-\frac{2 M}{r_{2}}\right)^{\frac{1}{2}}=v_{t}\left(r_{1}\right) \tag{E.197} \end{equation*}(E.197)vt(r2)=(12Mr2)12=vt(r1)
giving
v t ( r 1 ) = g t t ( r 1 ) v t ( r 1 ) = ( 1 2 M r 1 ) 1 ( 1 2 M r 2 ) (E.198) 1 2 v t r 1 = g t t r 1 v t r 1 = 1 2 M r 1 1 1 2 M r 2 (E.198)  1 2 v^(t)(r_(1))=g^(tt)(r_(1))v_(t)(r_(1))=(1-(2M)/(r_(1)))^(-1)(1-(2M)/(r_(2)))_((E.198) )^((1)/(2))v^{t}\left(r_{1}\right)=g^{t t}\left(r_{1}\right) v_{t}\left(r_{1}\right)=\left(1-\frac{2 M}{r_{1}}\right)^{-1}\left(1-\frac{2 M}{r_{2}}\right)_{\text {(E.198) }}^{\frac{1}{2}}vt(r1)=gtt(r1)vt(r1)=(12Mr1)1(12Mr2)(E.198) 12.
The dot product then yields
γ ( v rel ) = u v = g t t ( r 1 ) u t ( r 1 ) v t ( r 1 ) = ( 1 2 M r 1 ) ( 1 2 M r 1 ) 1 2 × ( 1 2 M r 1 ) 1 ( 1 2 M r 2 ) 1 2 (E.199) = ( 1 2 M / r 2 1 2 M / r 1 ) 1 2 γ v rel = u v = g t t r 1 u t r 1 v t r 1 = 1 2 M r 1 1 2 M r 1 1 2 × 1 2 M r 1 1 1 2 M r 2 1 2 (E.199) = 1 2 M / r 2 1 2 M / r 1 1 2 {:[-gamma(v_(rel))=u*v=g_(tt)(r_(1))u^(t)(r_(1))v^(t)(r_(1))],[=-(1-(2M)/(r_(1)))(1-(2M)/(r_(1)))^(-(1)/(2))],[ xx(1-(2M)/(r_(1)))^(-1)(1-(2M)/(r_(2)))^((1)/(2))],[(E.199)=-((1-2M//r_(2))/(1-2M//r_(1)))^((1)/(2))]:}\begin{align*} -\gamma\left(v_{\mathrm{rel}}\right)= & \boldsymbol{u} \cdot \boldsymbol{v}=g_{t t}\left(r_{1}\right) u^{t}\left(r_{1}\right) v^{t}\left(r_{1}\right) \\ = & -\left(1-\frac{2 M}{r_{1}}\right)\left(1-\frac{2 M}{r_{1}}\right)^{-\frac{1}{2}} \\ & \times\left(1-\frac{2 M}{r_{1}}\right)^{-1}\left(1-\frac{2 M}{r_{2}}\right)^{\frac{1}{2}} \\ = & -\left(\frac{1-2 M / r_{2}}{1-2 M / r_{1}}\right)^{\frac{1}{2}} \tag{E.199} \end{align*}γ(vrel)=uv=gtt(r1)ut(r1)vt(r1)=(12Mr1)(12Mr1)12×(12Mr1)1(12Mr2)12(E.199)=(12M/r212M/r1)12
Since γ ( v rel ) = ( 1 v rel 2 ) 1 2 γ v rel  = 1 v rel  2 1 2 gamma(v_("rel "))=(1-v_("rel ")^(2))^(-(1)/(2))\gamma\left(v_{\text {rel }}\right)=\left(1-v_{\text {rel }}^{2}\right)^{-\frac{1}{2}}γ(vrel )=(1vrel 2)12, the latter can be rearranged to access v rel v rel  v_("rel ")v_{\text {rel }}vrel .
(23.1) (a) We have d r = d θ = 0 d r = d θ = 0 dr=dtheta=0\mathrm{d} r=\mathrm{d} \theta=0dr=dθ=0, and so the increment of proper time taken along the orbit is
(E.200) d τ 2 = ( 1 2 M r ) d t 2 r 2 d ϕ 2 (E.200) d τ 2 = 1 2 M r d t 2 r 2 d ϕ 2 {:(E.200)dtau^(2)=(1-(2M)/(r))dt^(2)-r^(2)dphi^(2):}\begin{equation*} \mathrm{d} \tau^{2}=\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}-r^{2} \mathrm{~d} \phi^{2} \tag{E.200} \end{equation*}(E.200)dτ2=(12Mr)dt2r2 dϕ2
Substituting d ϕ 2 = v 2 r 2 d t 2 d ϕ 2 = v 2 r 2 d t 2 dphi^(2)=(v^(2))/(r^(2))dt^(2)\mathrm{d} \phi^{2}=\frac{v^{2}}{r^{2}} \mathrm{~d} t^{2}dϕ2=v2r2 dt2 we find
d τ = ± [ ( 1 2 M r ) v 2 ] 1 2 d t d τ = ± 1 2 M r v 2 1 2 d t dtau=+-[(1-(2M)/(r))-v^(2)]^((1)/(2))dt\mathrm{d} \tau= \pm\left[\left(1-\frac{2 M}{r}\right)-v^{2}\right]^{\frac{1}{2}} \mathrm{~d} tdτ=±[(12Mr)v2]12 dt
Everything in the square bracket is constant, and the integral d t d t intdt\int \mathrm{d} tdt over a period T T TTT yields d t = 2 π r / v d t = 2 π r / v intdt=2pi r//v\int \mathrm{d} t=2 \pi r / vdt=2πr/v, giving
(E.202) Δ τ = 2 π r v [ ( 1 2 M r ) v 2 ] 1 2 (E.202) Δ τ = 2 π r v 1 2 M r v 2 1 2 {:(E.202)Delta tau=(2pi r)/(v)[(1-(2M)/(r))-v^(2)]^((1)/(2)):}\begin{equation*} \Delta \tau=\frac{2 \pi r}{v}\left[\left(1-\frac{2 M}{r}\right)-v^{2}\right]^{\frac{1}{2}} \tag{E.202} \end{equation*}(E.202)Δτ=2πrv[(12Mr)v2]12
(b) Using the definition of L ~ L ~ tilde(L)\tilde{L}L~ and eqn 23.11 we find
( u t ) 2 = L ~ 2 / M r . u t 2 = L ~ 2 / M r . (u^(t))^(2)= tilde(L)^(2)//Mr.\left(u^{t}\right)^{2}=\tilde{L}^{2} / M r .(ut)2=L~2/Mr.
(E.203)
Now use this with the definition of E ~ E ~ tilde(E)\tilde{E}E~, combined with eqn 23.5 to obtain
(E.204) d τ = ( 1 3 M r ) 1 2 d t (E.204) d τ = 1 3 M r 1 2 d t {:(E.204)dtau=(1-(3M)/(r))^((1)/(2))dt:}\begin{equation*} \mathrm{d} \tau=\left(1-\frac{3 M}{r}\right)^{\frac{1}{2}} \mathrm{~d} t \tag{E.204} \end{equation*}(E.204)dτ=(13Mr)12 dt
Kepler's law tells us Δ t Δ t Delta t\Delta tΔt for an orbit is equal to 2 π ( r 3 / M ) 1 2 2 π r 3 / M 1 2 2pi(r^(3)//M)^((1)/(2))2 \pi\left(r^{3} / M\right)^{\frac{1}{2}}2π(r3/M)12, from which we obtain the answer. Kepler's law also implies that v = ( M / r ) 1 2 v = ( M / r ) 1 2 v=(M//r)^((1)/(2))v=(M / r)^{\frac{1}{2}}v=(M/r)12, which shows the answers from (a) and (b) are compatible.
(23.2) The method for solving problems of this type is explained in the answer to Exercise 23.1. Computing the proper time from the line element, we find Δ τ = Δ τ = Delta tau=\Delta \tau=Δτ= T ( 1 v 2 ) 1 2 T 1 v 2 1 2 T(1-v^(2))^((1)/(2))T\left(1-v^{2}\right)^{\frac{1}{2}}T(1v2)12, where T T TTT is the period in coordinate time. We then integrate the equation for v = e t R d ϕ / d t v = e t R d ϕ / d t v=e^(t)Rdphi//dtv=\mathrm{e}^{t} R \mathrm{~d} \phi / \mathrm{d} tv=etR dϕ/dt, to find
(E.205) T = ln ( 1 2 π R v ) 1 (E.205) T = ln 1 2 π R v 1 {:(E.205)T=ln (1-(2pi R)/(v))^(-1):}\begin{equation*} T=\ln \left(1-\frac{2 \pi R}{v}\right)^{-1} \tag{E.205} \end{equation*}(E.205)T=ln(12πRv)1
(23.3) A free particle follows a geodesic with constant E ~ = E ~ = tilde(E)=\tilde{E}=E~= A free particle follows a geodesic
( 1 2 M / r ) u t ( 1 2 M / r ) u t (1-2M//r)u^(t)(1-2 M / r) u^{t}(12M/r)ut. We therefore have
(E.206) d t d τ = ( 1 2 M r ) 1 E ~ = ( const. ) (E.206) d t d τ = 1 2 M r 1 E ~ = (  const.  ) {:(E.206)(dt)/((d)tau)=(1-(2M)/(r))^(-1) tilde(E)=(" const. "):}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} \tau}=\left(1-\frac{2 M}{r}\right)^{-1} \tilde{E}=(\text { const. }) \tag{E.206} \end{equation*}(E.206)dt dτ=(12Mr)1E~=( const. )
We conclude that t = c τ + d t = c τ + d t=c tau+dt=c \tau+dt=cτ+d, with c c ccc and d d ddd constant. Note that a similar argument applies to the ϕ ϕ phi\phiϕ variable Note that a similar argument applie
for motion at a constant value of θ θ theta\thetaθ.
(23.4) (a) We have u μ = ( C , 0 , 0 , ω ) u μ = ( C , 0 , 0 , ω ) u^(mu)=(C,0,0,omega)u^{\mu}=(C, 0,0, \omega)uμ=(C,0,0,ω) and, using u 2 = 1 u 2 = 1 u^(2)=-1\boldsymbol{u}^{2}=-1u2=1, we find
(E.207) C 2 = ( 1 + ω 2 r 2 1 2 M / r 0 ) (E.207) C 2 = 1 + ω 2 r 2 1 2 M / r 0 {:(E.207)C^(2)=((1+omega^(2)r^(2))/(1-2M//r_(0))):}\begin{equation*} C^{2}=\left(\frac{1+\omega^{2} r^{2}}{1-2 M / r_{0}}\right) \tag{E.207} \end{equation*}(E.207)C2=(1+ω2r212M/r0)
(b) Taking a covariant derivative yields
u u = ( Γ μ t t C 2 + Γ μ t ϕ C ω + Γ μ ϕ ϕ ω 2 ) e μ u u = Γ μ t t C 2 + Γ μ t ϕ C ω + Γ μ ϕ ϕ ω 2 e μ grad_(u)u=(Gamma^(mu)_(tt)C^(2)+Gamma^(mu)_(t phi)C omega+Gamma^(mu)_(phi phi)omega^(2))e_(mu)\boldsymbol{\nabla}_{u} \boldsymbol{u}=\left(\Gamma^{\mu}{ }_{t t} C^{2}+\Gamma^{\mu}{ }_{t \phi} C \omega+\Gamma^{\mu}{ }_{\phi \phi} \omega^{2}\right) \boldsymbol{e}_{\mu}uu=(ΓμttC2+ΓμtϕCω+Γμϕϕω2)eμ. (E.208) The only connection coefficients we need are Γ r t t = Γ r t t = Gamma^(r)_(tt)=\Gamma^{r}{ }_{t t}=Γrtt= M ( r 0 2 M ) / r 0 3 M r 0 2 M / r 0 3 M(r_(0)-2M)//r_(0)^(3)M\left(r_{0}-2 M\right) / r_{0}^{3}M(r02M)/r03, and Γ ϕ ϕ r = ( r 0 2 M ) Γ ϕ ϕ r = r 0 2 M Gamma_(phi phi)^(r)=-(r_(0)-2M)\Gamma_{\phi \phi}^{r}=-\left(r_{0}-2 M\right)Γϕϕr=(r02M), giving
(E.209) a = ( M r 0 2 + 3 M ω 2 r 0 ω 2 ) e r (E.209) a = M r 0 2 + 3 M ω 2 r 0 ω 2 e r {:(E.209)a=((M)/(r_(0)^(2))+3Momega^(2)-r_(0)omega^(2))e_(r):}\begin{equation*} \boldsymbol{a}=\left(\frac{M}{r_{0}^{2}}+3 M \omega^{2}-r_{0} \omega^{2}\right) \boldsymbol{e}_{r} \tag{E.209} \end{equation*}(E.209)a=(Mr02+3Mω2r0ω2)er
(c) The acceleration vanishes when
(E.210) M r 0 2 = ( r 0 3 M ) ω 2 (E.210) M r 0 2 = r 0 3 M ω 2 {:(E.210)(M)/(r_(0)^(2))=(r_(0)-3M)omega^(2):}\begin{equation*} \frac{M}{r_{0}^{2}}=\left(r_{0}-3 M\right) \omega^{2} \tag{E.210} \end{equation*}(E.210)Mr02=(r03M)ω2
For a particle far from the source of the field we need M / r 0 2 = r 0 ω 2 M / r 0 2 = r 0 ω 2 M//r_(0)^(2)=r_(0)omega^(2)M / r_{0}^{2}=r_{0} \omega^{2}M/r02=r0ω2, i.e. the gravitational acceleration supplies the centripetal acceleration required to keep the particle in orbit. When this is the case the particle follows a geodesic and we have no acceleration.
(23.5) (b) Referring to the figure in the chapter, we cut out a wedge δ δ delta\deltaδ such that
(E.211) ( 2 π δ ) R = 2 π r (E.211) ( 2 π δ ) R = 2 π r {:(E.211)(2pi-delta)R=2pi r:}\begin{equation*} (2 \pi-\delta) R=2 \pi r \tag{E.211} \end{equation*}(E.211)(2πδ)R=2πr
where r = R cos α r = R cos α r=R cos alphar=R \cos \alphar=Rcosα and tan α = d z / d r tan α = d z / d r tan alpha=dz//dr\tan \alpha=\mathrm{d} z / \mathrm{d} rtanα=dz/dr. Treating α α alpha\alphaα as a small angle, we have
(E.212) ( 2 π δ ) 2 π [ 1 1 2 ( d z d r ) 2 ] (E.212) ( 2 π δ ) 2 π 1 1 2 d z d r 2 {:(E.212)(2pi-delta)~~2pi[1-(1)/(2)(((d)z)/((d)r))^(2)]:}\begin{equation*} (2 \pi-\delta) \approx 2 \pi\left[1-\frac{1}{2}\left(\frac{\mathrm{~d} z}{\mathrm{~d} r}\right)^{2}\right] \tag{E.212} \end{equation*}(E.212)(2πδ)2π[112( dz dr)2]
Differentiating z ( r ) z ( r ) z(r)z(r)z(r) and inserting into this equation, we find
(E.213) δ 2 π M r 2 M 2 π M r (E.213) δ 2 π M r 2 M 2 π M r {:(E.213)delta~~(2pi M)/(r-2M)~~(2pi M)/(r):}\begin{equation*} \delta \approx \frac{2 \pi M}{r-2 M} \approx \frac{2 \pi M}{r} \tag{E.213} \end{equation*}(E.213)δ2πMr2M2πMr
where in the final term we have approximated r 2 M r 2 M r≫2Mr \gg 2 Mr2M. (d) One third of the effect is accounted from this contribution. As pointed out in Zee (2013), rubbersheet models showing a constant-time slice of the Schwarzschild geometry can be misleading in trying to understand the origin of gravitational effects in relativity.
(23.6) (a) We find
(E.214) u ϕ = [ E ~ 2 ( 1 2 M u ) 1 1 L ~ 2 u 2 ] 1 2 (E.214) u ϕ = E ~ 2 ( 1 2 M u ) 1 1 L ~ 2 u 2 1 2 {:(E.214)(del u)/(del phi)=∓[ tilde(E)^(2)(1-2Mu)^(-1)-1- tilde(L)^(2)u^(2)]^((1)/(2)):}\begin{equation*} \frac{\partial u}{\partial \phi}=\mp\left[\tilde{E}^{2}(1-2 M u)^{-1}-1-\tilde{L}^{2} u^{2}\right]^{\frac{1}{2}} \tag{E.214} \end{equation*}(E.214)uϕ=[E~2(12Mu)11L~2u2]12
(c) In the limit of small u c u c u_(c)u_{c}uc, we have u c M E ~ 2 / L ~ 2 u c M E ~ 2 / L ~ 2 u_(c)~~M tilde(E)^(2)// tilde(L)^(2)u_{c} \approx M \tilde{E}^{2} / \tilde{L}^{2}ucME~2/L~2. (d) Substituting, we obtain
(E.215) w + w + 1 M E ~ 2 L ~ 2 ( 1 u c + 4 M + 4 M w ) (E.215) w + w + 1 M E ~ 2 L ~ 2 1 u c + 4 M + 4 M w {:(E.215)w^('')+w+1~~(M tilde(E)^(2))/( tilde(L)^(2))*((1)/(u_(c))+4M+4Mw):}\begin{equation*} w^{\prime \prime}+w+1 \approx \frac{M \tilde{E}^{2}}{\tilde{L}^{2}} \cdot\left(\frac{1}{u_{\mathrm{c}}}+4 M+4 M w\right) \tag{E.215} \end{equation*}(E.215)w+w+1ME~2L~2(1uc+4M+4Mw)
or, in the limit of small u c u c u_(c)u_{c}uc
(E.216) w + ( 1 4 M u c ) w 0 (E.216) w + 1 4 M u c w 0 {:(E.216)w^('')+(1-4Mu_(c))w~~0:}\begin{equation*} w^{\prime \prime}+\left(1-4 M u_{\mathrm{c}}\right) w \approx 0 \tag{E.216} \end{equation*}(E.216)w+(14Muc)w0
This is a simple harmonic oscillator equation with characteristic frequency ω 0 2 = 1 4 M u c ω 0 2 = 1 4 M u c omega_(0)^(2)=1-4Mu_(c)\omega_{0}^{2}=1-4 M u_{\mathrm{c}}ω02=14Muc.
(e) Using the hint we have
(E.217) ω 0 Δ ϕ = 2 π (E.217) ω 0 Δ ϕ = 2 π {:(E.217)omega_(0)Delta phi=2pi:}\begin{equation*} \omega_{0} \Delta \phi=2 \pi \tag{E.217} \end{equation*}(E.217)ω0Δϕ=2π
which can be rearranged and expanded to find
(E.218) Δ ϕ 2 π + 4 π M r c (E.218) Δ ϕ 2 π + 4 π M r c {:(E.218)Delta phi~~2pi+(4pi M)/(r_(c)):}\begin{equation*} \Delta \phi \approx 2 \pi+\frac{4 \pi M}{r_{\mathrm{c}}} \tag{E.218} \end{equation*}(E.218)Δϕ2π+4πMrc
Note that the latter is 2 / 3 2 / 3 2//32 / 32/3 of the total perihelion shift. (24.3) (a) From Fig. 24.8, we have
(E.219) θ i D s = θ s D s + α D s . (E.219) θ i D s = θ s D s + α D s . {:(E.219)theta_(i)D_(s)=theta_(s)D_(s)+alphaD_(ℓs).:}\begin{equation*} \theta_{\mathrm{i}} D_{\mathrm{s}}=\theta_{\mathrm{s}} D_{\mathrm{s}}+\alpha D_{\ell \mathrm{s}} . \tag{E.219} \end{equation*}(E.219)θiDs=θsDs+αDs.
However, we also have that α = 4 M / b α = 4 M / b alpha=4M//b\alpha=4 M / bα=4M/b is the deflection angle, where b = θ i D b = θ i D b=theta_(i)D_(ℓ)b=\theta_{\mathrm{i}} D_{\ell}b=θiD, so that
(E.220) θ i = θ s + 4 M D s θ i D D s (E.220) θ i = θ s + 4 M D s θ i D D s {:(E.220)theta_(i)=theta_(s)+(4MD_(ℓs))/(theta_(i)D_(ℓ)D_(s)):}\begin{equation*} \theta_{\mathrm{i}}=\theta_{\mathrm{s}}+\frac{4 M D_{\ell \mathrm{s}}}{\theta_{\mathrm{i}} D_{\ell} D_{\mathrm{s}}} \tag{E.220} \end{equation*}(E.220)θi=θs+4MDsθiDDs
The answer then follows using the definition of θ E θ E theta_(E)\theta_{\mathrm{E}}θE.
(b) We have that D s D s D_(ℓs)D_{\ell \mathrm{s}}Ds and D s D s D_(s)D_{\mathrm{s}}Ds are both far larger than D D D_(ℓ)D_{\ell}D, which gives θ E 2 4 M / D θ E 2 4 M / D theta_(E)^(2)~~4M//D_(ℓ)\theta_{\mathrm{E}}^{2} \approx 4 M / D_{\ell}θE24M/D. For the case where the source, lens, and observer are collinear, we have θ s = 0 θ s = 0 theta_(s)=0\theta_{\mathrm{s}}=0θs=0 and θ i = θ E θ i = θ E theta_(i)=theta_(E)\theta_{\mathrm{i}}=\theta_{\mathrm{E}}θi=θE. The image is then a ring with this angular radius.
(24.4) (a) Write d χ 2 = d r 2 1 k r 2 d χ 2 = d r 2 1 k r 2 dchi^(2)=(dr^(2))/(1-kr^(2))\mathrm{d} \chi^{2}=\frac{\mathrm{d} r^{2}}{1-k r^{2}}dχ2=dr21kr2. For a radial ray d ϕ = d θ = 0 d ϕ = d θ = 0 dphi=dtheta=0\mathrm{d} \phi=\mathrm{d} \theta=0dϕ=dθ=0 and the result follows.
(b) Metric coefficients are independent of χ χ chi\chiχ along the ray and so we have a Killing vector e χ e χ e_(chi)\boldsymbol{e}_{\chi}eχ and so u χ = C u χ = C u_(chi)=Cu_{\chi}=Cuχ=C is conserved along the ray. Raising the index using g χ χ g χ χ g^(chi chi)g^{\chi \chi}gχχ gives
u χ = C / a ( t ) 2 u χ = C / a ( t ) 2 u^(chi)=C//a(t)^(2)u^{\chi}=C / a(t)^{2}uχ=C/a(t)2
(E.221)
Since u χ = d χ d λ u χ = d χ d λ u^(chi)=(dchi)/(dlambda)u^{\chi}=\frac{\mathrm{d} \chi}{\mathrm{d} \mathrm{\lambda}}uχ=dχdλ, where λ λ lambda\lambdaλ is the affine parameter we have
d λ = C 1 a ( t ) 2 d χ = 1 C a ( t ) 2 d r ( 1 k r 2 ) 1 2 d λ = C 1 a ( t ) 2 d χ = 1 C a ( t ) 2 d r 1 k r 2 1 2 dlambda=C^(-1)a(t)^(2)dchi=(1)/(C)(a(t)^(2)(d)r)/((1-kr^(2))^((1)/(2)))\mathrm{d} \lambda=C^{-1} a(t)^{2} \mathrm{~d} \chi=\frac{1}{C} \frac{a(t)^{2} \mathrm{~d} r}{\left(1-k r^{2}\right)^{\frac{1}{2}}}dλ=C1a(t)2 dχ=1Ca(t)2 dr(1kr2)12
(25.1) (a) Substituting yields d s 2 = ( 1 v 2 ) d T 2 + [ ( 1 v 2 ) 1 R 2 ( 1 v 2 ) ] d r 2 d s 2 = 1 v 2 d T 2 + 1 v 2 1 R 2 1 v 2 d r 2 ds^(2)=-(1-v^(2))dT^(2)+[(1-v^(2))^(-1)-R^(2)(1-v^(2))]dr^(2)\mathrm{d} s^{2}=-\left(1-v^{2}\right) \mathrm{d} T^{2}+\left[\left(1-v^{2}\right)^{-1}-R^{2}\left(1-v^{2}\right)\right] \mathrm{d} r^{2}ds2=(1v2)dT2+[(1v2)1R2(1v2)]dr2
(E.223) + 2 ( 1 v 2 ) R d r d T + r 2 d Ω 2 . (E.223) + 2 1 v 2 R d r d T + r 2 d Ω 2 . {:(E.223)+2(1-v^(2))RdrdT+r^(2)dOmega^(2).:}\begin{equation*} +2\left(1-v^{2}\right) R \mathrm{~d} r \mathrm{~d} T+r^{2} \mathrm{~d} \Omega^{2} . \tag{E.223} \end{equation*}(E.223)+2(1v2)R dr dT+r2 dΩ2.
(b) Setting the square bracket to equal unity yields R = v ( 1 v 2 ) 1 R = v 1 v 2 1 R=v(1-v^(2))^(-1)R=v\left(1-v^{2}\right)^{-1}R=v(1v2)1.
(d) A slice of space at a constant value of T T TTT is flat (e) The velocity d r / d τ d r / d τ dr//dtau\mathrm{d} r / \mathrm{d} \taudr/dτ is unchanged from the Schwarzschild value u r = ( 2 M / r ) 1 2 u r = ( 2 M / r ) 1 2 u^(r)=-(2M//r)^((1)/(2))u^{r}=-(2 M / r)^{\frac{1}{2}}ur=(2M/r)12. Making the substitution and rearranging we find that we also have
(E.224) d r d T = ( 2 M r ) 1 2 = | v | . (E.224) d r d T = 2 M r 1 2 = | v | . {:(E.224)(dr)/((d)T)=-((2M)/(r))^((1)/(2))=-|v|.:}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} T}=-\left(\frac{2 M}{r}\right)^{\frac{1}{2}}=-|v| . \tag{E.224} \end{equation*}(E.224)dr dT=(2Mr)12=|v|.
This is unity at the event horizon, diverging as r 0 r 0 r rarr0r \rightarrow 0r0. (f) Setting d s 2 = 0 d s 2 = 0 ds^(2)=0\mathrm{d} s^{2}=0ds2=0 we find that
(E.225) d r d T = ± 1 | v | (E.225) d r d T = ± 1 | v | {:(E.225)(dr)/((d)T)=+-1-|v|:}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} T}= \pm 1-|v| \tag{E.225} \end{equation*}(E.225)dr dT=±1|v|
(g) We have a ratio Q Q QQQ of the coordinate speed of the in-falling observer ( | v | ) ( | v | ) (-|v|)(-|v|)(|v|) to in-falling light ( 1 | v | ) ( 1 | v | ) (-1-|v|)(-1-|v|)(1|v|) of
(E.226) Q = | v | 1 + | v | = ( 2 M r ) 1 2 [ 1 + ( 2 M r ) 1 2 ] 1 < 1 (E.226) Q = | v | 1 + | v | = 2 M r 1 2 1 + 2 M r 1 2 1 < 1 {:(E.226)Q=(|v|)/(1+|v|)=((2M)/(r))^((1)/(2))[1+((2M)/(r))^((1)/(2))]^(-1) < 1:}\begin{equation*} Q=\frac{|v|}{1+|v|}=\left(\frac{2 M}{r}\right)^{\frac{1}{2}}\left[1+\left(\frac{2 M}{r}\right)^{\frac{1}{2}}\right]^{-1}<1 \tag{E.226} \end{equation*}(E.226)Q=|v|1+|v|=(2Mr)12[1+(2Mr)12]1<1
and so the speed appears subluminal.
(25.2) (a) We first want to evaluate the interval between two events at fixed r r rrr, measured by falling clocks. The interval is measured by one clock that falls through r r rrr and a second clock that falls through r r rrr an interval d t d t dt\mathrm{d} tdt later. Each clock requires the same interval in t t ttt to fall to r r rrr, so the second clock must have been dropped d t d t dt\mathrm{d} tdt seconds after the first. The difference in d t ˘ d t ˘ dt^(˘)\mathrm{d} \breve{t}dt˘ registered between the arrival of the clocks is simply the difference d t d t dt\mathrm{d} tdt between the clocks being dropped and so d t ˘ = d t d t ˘ = d t dt^(˘)=dt\mathrm{d} \breve{t}=\mathrm{d} tdt˘=dt.
(b) We are in the same situation as examined in Chapter 22 for radially falling observers. We saw there that d r / d τ = ( 2 M / r ) 1 2 d r / d τ = ( 2 M / r ) 1 2 dr//dtau=-(2M//r)^((1)/(2))\mathrm{d} r / \mathrm{d} \tau=-(2 M / r)^{\frac{1}{2}}dr/dτ=(2M/r)12 and
(E.227) d r d t = ( 1 2 M r ) ( 2 M r ) 1 2 (E.227) d r d t = 1 2 M r 2 M r 1 2 {:(E.227)(dr)/((d)t)=-(1-(2M)/(r))((2M)/(r))^((1)/(2)):}\begin{equation*} \frac{\mathrm{d} r}{\mathrm{~d} t}=-\left(1-\frac{2 M}{r}\right)\left(\frac{2 M}{r}\right)^{\frac{1}{2}} \tag{E.227} \end{equation*}(E.227)dr dt=(12Mr)(2Mr)12
(c) Write d t d = d r / ( d r / d t ) d t d = d r / ( d r / d t ) dt_(d)=-dr//(dr//dt)\mathrm{d} t_{\mathrm{d}}=-\mathrm{d} r /(\mathrm{d} r / \mathrm{d} t)dtd=dr/(dr/dt) and substitute from the previous equation.
(d) Use eqn 22.31 to compute d τ d τ dtau\mathrm{d} \taudτ via the integral
(E.228) Δ τ = r A r B d r ( r 2 M ) 1 2 (E.228) Δ τ = r A r B d r r 2 M 1 2 {:(E.228)Delta tau=-int_(r_(A))^(r_(B))dr((r)/(2M))^((1)/(2)):}\begin{equation*} \Delta \tau=-\int_{r_{\mathrm{A}}}^{r_{\mathrm{B}}} \mathrm{~d} r\left(\frac{r}{2 M}\right)^{\frac{1}{2}} \tag{E.228} \end{equation*}(E.228)Δτ=rArB dr(r2M)12
(e) The total time delay is the sum of (i) the delay in drop time and (ii) the delay in proper time to reach their different radial positions at a time t t ttt.
(f) We evaluate
t ˘ r = ( d τ + d t d ) d r = ( r 2 M ) 1 2 [ 1 + ( 1 2 M r ) 1 ] = ( 2 M r ) 1 2 ( 1 2 M r ) 1 t ˘ r = d τ + d t d d r = r 2 M 1 2 1 + 1 2 M r 1 = 2 M r 1 2 1 2 M r 1 {:[(del(t^(˘)))/(del r)=((dtau+dt_(d)))/(dr)],[=((r)/(2M))^((1)/(2))[-1+(1-(2M)/(r))^(-1)]],[=((2M)/(r))^((1)/(2))(1-(2M)/(r))^(-1)]:}\begin{aligned} \frac{\partial \breve{t}}{\partial r} & =\frac{\left(\mathrm{d} \tau+\mathrm{d} t_{\mathrm{d}}\right)}{\mathrm{d} r} \\ & =\left(\frac{r}{2 M}\right)^{\frac{1}{2}}\left[-1+\left(1-\frac{2 M}{r}\right)^{-1}\right] \\ & =\left(\frac{2 M}{r}\right)^{\frac{1}{2}}\left(1-\frac{2 M}{r}\right)^{-1} \end{aligned}t˘r=(dτ+dtd)dr=(r2M)12[1+(12Mr)1]=(2Mr)12(12Mr)1
(25.3) From the previous problem we have that
(E.230) d t ˘ = d t + ( 2 M / r ) 1 2 1 2 M / r d r (E.230) d t ˘ = d t + ( 2 M / r ) 1 2 1 2 M / r d r {:(E.230)dt^(˘)=dt+((2M//r)^((1)/(2)))/(1-2M//r)dr:}\begin{equation*} \mathrm{d} \breve{t}=\mathrm{d} t+\frac{(2 M / r)^{\frac{1}{2}}}{1-2 M / r} \mathrm{~d} r \tag{E.230} \end{equation*}(E.230)dt˘=dt+(2M/r)1212M/r dr
Substituting for d t d t dt\mathrm{d} tdt in the Schwarzschild metric gives the desired answer.
(25.4) (a) The stationary observer has only one non-zero velocity component u t u t u^(t)u^{t}ut, determined by the equation g t t ( u t ) 2 = 1 g t t u t 2 = 1 g_(tt)(u^(t))^(2)=-1g_{t t}\left(u^{t}\right)^{2}=-1gtt(ut)2=1, giving u t = ( 1 2 M / r ) 1 2 u t = ( 1 2 M / r ) 1 2 u^(t)=(1-2M//r)^(-(1)/(2))u^{t}=(1-2 M / r)^{-\frac{1}{2}}ut=(12M/r)12. We can therefore write
(E.231) ( 1 v 0 2 ) 1 2 = g t t u t v t (E.231) 1 v 0 2 1 2 = g t t u t v t {:(E.231)-(1-v_(0)^(2))^(-(1)/(2))=g_(tt)u^(t)v^(t):}\begin{equation*} -\left(1-v_{0}^{2}\right)^{-\frac{1}{2}}=g_{t t} u^{t} v^{t} \tag{E.231} \end{equation*}(E.231)(1v02)12=gttutvt
or
(E.232) v t = ( 1 v 0 2 ) 1 2 ( 1 2 M / r ) 1 2 (E.232) v t = 1 v 0 2 1 2 ( 1 2 M / r ) 1 2 {:(E.232)v^(t)=(1-v_(0)^(2))^(-(1)/(2))(1-2M//r)^(-(1)/(2)):}\begin{equation*} v^{t}=\left(1-v_{0}^{2}\right)^{-\frac{1}{2}}(1-2 M / r)^{-\frac{1}{2}} \tag{E.232} \end{equation*}(E.232)vt=(1v02)12(12M/r)12
The only other non-zero component of v μ v μ v^(mu)v^{\mu}vμ is v ϕ v ϕ v^(phi)v^{\phi}vϕ and so, using g ( v , v ) = 1 g ( v , v ) = 1 g(v,v)=-1\boldsymbol{g}(\boldsymbol{v}, \boldsymbol{v})=-1g(v,v)=1, we have
(E.233) ( 1 v 0 2 ) 1 + r 2 ( v ϕ ) 2 = 1 (E.234) v ϕ = ( v 0 2 / r 0 2 1 v 0 2 ) 1 2 (E.233) 1 v 0 2 1 + r 2 v ϕ 2 = 1 (E.234) v ϕ = v 0 2 / r 0 2 1 v 0 2 1 2 {:[(E.233)-(1-v_(0)^(2))^(-1)+r^(2)(v^(phi))^(2)=-1],[(E.234)v^(phi)=((v_(0)^(2)//r_(0)^(2))/(1-v_(0)^(2)))^((1)/(2))]:}\begin{gather*} -\left(1-v_{0}^{2}\right)^{-1}+r^{2}\left(v^{\phi}\right)^{2}=-1 \tag{E.233}\\ v^{\phi}=\left(\frac{v_{0}^{2} / r_{0}^{2}}{1-v_{0}^{2}}\right)^{\frac{1}{2}} \tag{E.234} \end{gather*}(E.233)(1v02)1+r2(vϕ)2=1(E.234)vϕ=(v02/r021v02)12
or
or
(b) We compute
(E.235) E ~ = ( 1 2 M / r ) 1 2 ( 1 v 0 2 ) 1 2 (E.235) E ~ = ( 1 2 M / r ) 1 2 1 v 0 2 1 2 {:(E.235) tilde(E)=(1-2M//r)^((1)/(2))(1-v_(0)^(2))^(-(1)/(2)):}\begin{equation*} \tilde{E}=(1-2 M / r)^{\frac{1}{2}}\left(1-v_{0}^{2}\right)^{-\frac{1}{2}} \tag{E.235} \end{equation*}(E.235)E~=(12M/r)12(1v02)12
and
(E.236) L ~ = v 0 r ( 1 v 0 2 ) 1 2 (E.236) L ~ = v 0 r 1 v 0 2 1 2 {:(E.236) tilde(L)=(v_(0)r)/((1-v_(0)^(2))^((1)/(2))):}\begin{equation*} \tilde{L}=\frac{v_{0} r}{\left(1-v_{0}^{2}\right)^{\frac{1}{2}}} \tag{E.236} \end{equation*}(E.236)L~=v0r(1v02)12
(c) Plugging in, we find that for r = 4 M r = 4 M r=4Mr=4 Mr=4M we have E ~ = [ 2 ( 1 v 0 2 ) ] 1 2 E ~ = 2 1 v 0 2 1 2 tilde(E)=[2(1-v_(0)^(2))]^(-(1)/(2))\tilde{E}=\left[2\left(1-v_{0}^{2}\right)\right]^{-\frac{1}{2}}E~=[2(1v02)]12 and therefore
(E.237) E = v 0 2 1 2 2 2 v 0 2 . (E.237) E = v 0 2 1 2 2 2 v 0 2 . {:(E.237)E=(v_(0)^(2)-(1)/(2))/(2-2v_(0)^(2)).:}\begin{equation*} \mathcal{E}=\frac{v_{0}^{2}-\frac{1}{2}}{2-2 v_{0}^{2}} . \tag{E.237} \end{equation*}(E.237)E=v021222v02.
We also have L ~ = 4 M v 0 ( 1 v 0 2 ) 1 2 L ~ = 4 M v 0 1 v 0 2 1 2 tilde(L)=(4Mv_(0))/((1-v_(0)^(2))^((1)/(2)))\tilde{L}=\frac{4 M v_{0}}{\left(1-v_{0}^{2}\right)^{\frac{1}{2}}}L~=4Mv0(1v02)12 and
(E.238) V eff ( r ) = v 0 2 1 2 2 2 v 0 2 . (E.238) V eff ( r ) = v 0 2 1 2 2 2 v 0 2 . {:(E.238)V_(eff)(r)=(v_(0)^(2)-(1)/(2))/(2-2v_(0)^(2)).:}\begin{equation*} V_{\mathrm{eff}}(r)=\frac{v_{0}^{2}-\frac{1}{2}}{2-2 v_{0}^{2}} . \tag{E.238} \end{equation*}(E.238)Veff(r)=v021222v02.
This makes sense if the particle is deflected at r r rrr, since we must have E = V eff ( r ) E = V eff  ( r ) E=V_("eff ")(r)\mathcal{E}=V_{\text {eff }}(r)E=Veff (r) in order for the motion to be tangential at the deflection point. A value of v 0 = 1 / 2 v 0 = 1 / 2 v_(0)=1//sqrt2v_{0}=1 / \sqrt{2}v0=1/2 gives E = V eff = 0 E = V eff  = 0 E=V_("eff ")=0\mathcal{E}=V_{\text {eff }}=0E=Veff =0 and also L ~ / M = 4 M L ~ / M = 4 M tilde(L)//M=4M\tilde{L} / M=4 ML~/M=4M. From Fig. 23.1 we see that the maximum value of the L ~ / M = 4 M L ~ / M = 4 M tilde(L)//M=4M\tilde{L} / M=4 ML~/M=4M curve is V eff = 0 V eff  = 0 V_("eff ")=0V_{\text {eff }}=0Veff =0, occurring exactly at r = 4 M r = 4 M r=4Mr=4 Mr=4M. We conclude that the particle is, only just, deflected from this point. In fact, v 0 = 1 / 2 v 0 = 1 / 2 v_(0)=1//sqrt2v_{0}=1 / \sqrt{2}v0=1/2 represents the minimum value of the relative velocity that allows the particle to escape the hole from this point.
(26.4) (a) Note that e t = ( / t ) μ e t = ( / t ) μ e_(t)=(del//del t)^(mu)\boldsymbol{e}_{t}=(\partial / \partial t)^{\mu}et=(/t)μ is a Killing vector, and so u t u t u_(t)u_{t}ut is a constant of the motion, just as we've had previously. The constant can be rewritten using the metric components as
(E.239) E ~ = g t t u t = x 2 d t d λ (E.239) E ~ = g t t u t = x 2 d t d λ {:(E.239)- tilde(E)=g_(tt)u^(t)=-x^(2)((d)t)/((d)lambda):}\begin{equation*} -\tilde{E}=g_{t t} u^{t}=-x^{2} \frac{\mathrm{~d} t}{\mathrm{~d} \lambda} \tag{E.239} \end{equation*}(E.239)E~=gttut=x2 dt dλ
which is also a constant of the motion.
(b) Substitute for x x xxx and t t ttt, set u = u = u=u=u= const. and integrate to get the expression for the affine parameter for outgoing geodesics.
(27.4) Along with the world-line component x 1 ( τ ) = x 0 x 1 ( τ ) = x 0 x^(1)(tau)=x_(0)x^{1}(\tau)=x_{0}x1(τ)=x0, we use u 2 = 1 u 2 = 1 u^(2)=-1\boldsymbol{u}^{2}=-1u2=1 to compute g t t ( u t ) 2 = 1 g t t u t 2 = 1 g_(tt)(u^(t))^(2)=-1g_{t t}\left(u^{t}\right)^{2}=-1gtt(ut)2=1 and find that the velocity has components u μ = ( 1 / x 0 , 0 ) u μ = 1 / x 0 , 0 u^(mu)=(1//x_(0),0)u^{\mu}=\left(1 / x_{0}, 0\right)uμ=(1/x0,0). Taking a covariant derivative yields an acceleration with non-zero components
(E.240) a μ = u t u μ t + Γ t t μ ( u t ) 2 (E.240) a μ = u t u μ t + Γ t t μ u t 2 {:(E.240)a^(mu)=u^(t)(delu^(mu))/(del t)+Gamma_(tt)^(mu)(u^(t))^(2):}\begin{equation*} a^{\mu}=u^{t} \frac{\partial u^{\mu}}{\partial t}+\Gamma_{t t}^{\mu}\left(u^{t}\right)^{2} \tag{E.240} \end{equation*}(E.240)aμ=utuμt+Γttμ(ut)2
Using the connection coefficients from Exercise 9.4, we conclude a μ = ( 0 , 1 / x 0 ) a μ = 0 , 1 / x 0 a^(mu)=(0,1//x_(0))a^{\mu}=\left(0,1 / x_{0}\right)aμ=(0,1/x0). The proper acceleration α α alpha\alphaα has the property α 2 = a 2 α 2 = a 2 alpha^(2)=a^(2)\alpha^{2}=\boldsymbol{a}^{2}α2=a2, so we have α = 1 / x 0 α = 1 / x 0 alpha=1//x_(0)\alpha=1 / x_{0}α=1/x0 too.
27.5) (a) Identify τ τ tau\tauτ and the observer's proper time and γ = γ = gamma=\gamma=γ= ( 1 v 2 ) 1 2 1 v 2 1 2 (1-v^(2))^(-(1)/(2))\left(1-v^{2}\right)^{-\frac{1}{2}}(1v2)12 and the result follows.
(b) Substituting into the exponential we obtain
(E.241) A = exp [ i Ω γ ( 1 v ) τ ] (E.241) A = exp [ i Ω γ ( 1 v ) τ ] {:(E.241)A=exp[i Omega gamma(1-v)tau]:}\begin{equation*} A=\exp [i \Omega \gamma(1-v) \tau] \tag{E.241} \end{equation*}(E.241)A=exp[iΩγ(1v)τ]
which can be rewritten in terms of an effective frequency as
(E.242) Ω = Ω ( 1 v 1 + v ) 1 2 (E.242) Ω = Ω 1 v 1 + v 1 2 {:(E.242)Omega^(')=Omega((1-v)/(1+v))^((1)/(2)):}\begin{equation*} \Omega^{\prime}=\Omega\left(\frac{1-v}{1+v}\right)^{\frac{1}{2}} \tag{E.242} \end{equation*}(E.242)Ω=Ω(1v1+v)12
It's worth noting that the instantaneous frequency can be extracted by operating on the phase with i / τ i / τ -idel//del tau-\mathrm{i} \partial / \partial \taui/τ. (c) Repeating we obtain
(E.243) A = exp [ i Ω g ( e g τ ) ] (E.243) A = exp i Ω g e g τ {:(E.243)A=exp[-i(Omega )/(g)(e^(-g tau))]:}\begin{equation*} A=\exp \left[-\mathrm{i} \frac{\Omega}{g}\left(\mathrm{e}^{-g \tau}\right)\right] \tag{E.243} \end{equation*}(E.243)A=exp[iΩg(egτ)]
Acting on the phase with i / τ i / τ -idel//del tau-\mathrm{i} \partial / \partial \taui/τ yields up the frequency
(E.244) ω ( τ ) = Ω e g τ (E.244) ω ( τ ) = Ω e g τ {:(E.244)omega(tau)=Omegae^(-g tau):}\begin{equation*} \omega(\tau)=\Omega \mathrm{e}^{-g \tau} \tag{E.244} \end{equation*}(E.244)ω(τ)=Ωegτ
(28.1) The area A A AAA is given by
A = d A = g θ θ g ϕ ϕ d θ d ϕ = r S 2 sin θ d θ d ϕ (E.245) = 4 π r S 2 A = d A = g θ θ g ϕ ϕ d θ d ϕ = r S 2 sin θ d θ d ϕ (E.245) = 4 π r S 2 {:[A=intdA=intsqrt(g_(theta theta)g_(phi phi))dthetadphi],[=intr_(S)^(2)sin thetadthetadphi],[(E.245)=4pir_(S)^(2)]:}\begin{align*} A=\int \mathrm{d} A & =\int \sqrt{g_{\theta \theta} g_{\phi \phi}} \mathrm{d} \theta \mathrm{~d} \phi \\ & =\int r_{\mathrm{S}}^{2} \sin \theta \mathrm{~d} \theta \mathrm{~d} \phi \\ & =4 \pi r_{\mathrm{S}}^{2} \tag{E.245} \end{align*}A=dA=gθθgϕϕdθ dϕ=rS2sinθ dθ dϕ(E.245)=4πrS2
Since r S = 2 M r S = 2 M r_(S)=2Mr_{\mathrm{S}}=2 MrS=2M, we have A = 16 π M 2 A = 16 π M 2 A=16 piM^(2)A=16 \pi M^{2}A=16πM2.
(28.2) Use M = / ( 8 π k B T ) , A = 16 π M 2 , S = A k B / ( 4 ) M = / 8 π k B T , A = 16 π M 2 , S = A k B / ( 4 ) M=ℏ//(8pik_(B)T),A=16 piM^(2),S=Ak_(B)//(4ℏ)M=\hbar /\left(8 \pi k_{\mathrm{B}} T\right), A=16 \pi M^{2}, S=A k_{\mathrm{B}} /(4 \hbar)M=/(8πkBT),A=16πM2,S=AkB/(4), and the result follows. The negative sign results from the fact that when a Schwarzschild black hole accretes matter, its energy and mass increases, but its temperature decreases. Similarly, when the black hole evaporates, it radiates energy, its mass decreases, and its temperature increases. Hence the negative heat capacity.
(28.3) (a) The required metric components, in matrix form for the coordinates ( V , r ) ( V , r ) (V,r)(V, r)(V,r), are
(E.246) g μ ν = ( 1 + 2 M r 1 1 0 ) (E.246) g μ ν = 1 + 2 M r 1 1 0 {:(E.246)g_(mu nu)=([-1+(2M)/(r),1],[1,0]):}g_{\mu \nu}=\left(\begin{array}{rr} -1+\frac{2 M}{r} & 1 \tag{E.246}\\ 1 & 0 \end{array}\right)(E.246)gμν=(1+2Mr110)
Acting on ξ μ ξ μ xi^(mu)\xi^{\mu}ξμ with this matrix gives ξ ν ξ ν xi_(nu)\xi_{\nu}ξν.
(c) The motivation for the acrobatics in this problem is to use ξ μ = ( 1 , 0 , 0 , 0 ) ξ μ = ( 1 , 0 , 0 , 0 ) xi^(mu)=(1,0,0,0)\xi^{\mu}=(1,0,0,0)ξμ=(1,0,0,0) to simplify the expressions. Start by choosing σ = V σ = V sigma=V\sigma=Vσ=V in ξ μ ξ μ ; σ = κ ξ σ ξ μ ξ μ ; σ = κ ξ σ -xi^(mu)xi_(mu)^(;sigma)=kappaxi^(sigma)-\xi^{\mu} \xi_{\mu}^{; \sigma}=\kappa \xi^{\sigma}ξμξμ;σ=κξσ and then note that only the μ = V μ = V mu=V\mu=Vμ=V component survives in the contraction to obtain ξ V ; V = κ ξ V ; V = κ xi_(V)^(;V)=-kappa\xi_{V}{ }^{; V}=-\kappaξV;V=κ. There is a rule for dealing more directly with this equation for the diverdealing more directly with this equation for fie divergence that
can write
Then using
(E.247) ξ V ; σ g σ V = κ (E.247) ξ V ; σ g σ V = κ {:(E.247)xi_(V;sigma)g^(sigma V)=-kappa:}\begin{equation*} \xi_{V ; \sigma} g^{\sigma V}=-\kappa \tag{E.247} \end{equation*}(E.247)ξV;σgσV=κ
g μ ν = ( 0 1 1 1 2 M r ) , g μ ν = 0 1 1 1 2 M r , g^(mu nu)=([0,1],[1,1-(2M)/(r)]),g^{\mu \nu}=\left(\begin{array}{cc} 0 & 1 \\ 1 & 1-\frac{2 M}{r} \end{array}\right),gμν=(01112Mr),
(E.248)
we obtain the answer.
(28.4) (d) The angle ψ ψ psi\psiψ repeats after 2 π 2 π 2pi2 \pi2π radians so that ψ = ψ = psi=\psi=ψ= ψ + 2 π n ψ + 2 π n psi+2pi n\psi+2 \pi nψ+2πn, with n n nnn an integer. Since ψ = t E / 4 M ψ = t E / 4 M psi=t_(E)//4M\psi=t_{\mathrm{E}} / 4 Mψ=tE/4M, this gives the repeat period in imaginary time as Δ t E = 8 π M Δ t E = 8 π M Deltat_(E)=8pi M\Delta t_{\mathrm{E}}=8 \pi MΔtE=8πM, corresponding to a temperature
(E.249) k B T = c 3 8 π G M (E.249) k B T = c 3 8 π G M {:(E.249)k_(B)T=(ℏc^(3))/(8pi GM):}\begin{equation*} k_{\mathrm{B}} T=\frac{\hbar c^{3}}{8 \pi G M} \tag{E.249} \end{equation*}(E.249)kBT=c38πGM
(29.6) We want to compute
( g t t g t ϕ g ϕ t g ϕ ϕ ) 1 = 1 D ( g ϕ ϕ g t ϕ g ϕ t g t t ) g t t      g t ϕ g ϕ t      g ϕ ϕ 1 = 1 D g ϕ ϕ g t ϕ g ϕ t g t t ([g_(tt),g_(t phi)],[g_(phi t),g_(phi phi)])^(-1)=(1)/(D)([g_(phi phi),-g_(t phi)],[-g_(phi t),g_(tt)])\left(\begin{array}{ll} g_{t t} & g_{t \phi} \\ g_{\phi t} & g_{\phi \phi} \end{array}\right)^{-1}=\frac{1}{D}\left(\begin{array}{cc} g_{\phi \phi} & -g_{t \phi} \\ -g_{\phi t} & g_{t t} \end{array}\right)(gttgtϕgϕtgϕϕ)1=1D(gϕϕgtϕgϕtgtt)
(E.250)
where D D DDD is the determinant of the matrix in the question. Some algebra reveals D = Δ sin 2 θ D = Δ sin 2 θ D=-Deltasin^(2)thetaD=-\Delta \sin ^{2} \thetaD=Δsin2θ, from which the desired equations follow straightforwardly.
(29.8) Using d τ 2 = d s 2 d τ 2 = d s 2 dtau^(2)=-ds^(2)\mathrm{d} \tau^{2}=-\mathrm{d} s^{2}dτ2=ds2, we find
d τ 2 = [ ( 1 2 M R ) + 4 M a R 2 v ( 1 + a 2 R 2 + 2 M a 2 R 3 ) v 2 ] d t 2 . d τ 2 = 1 2 M R + 4 M a R 2 v 1 + a 2 R 2 + 2 M a 2 R 3 v 2 d t 2 . {:[dtau^(2)=[(1-(2M)/(R))+(4Ma)/(R^(2))*v:}],[{:-(1+(a^(2))/(R^(2))+(2Ma^(2))/(R^(3)))v^(2)]dt^(2).]:}\begin{align*} \mathrm{d} \tau^{2}= & {\left[\left(1-\frac{2 M}{R}\right)+\frac{4 M a}{R^{2}} \cdot v\right.} \\ & \left.-\left(1+\frac{a^{2}}{R^{2}}+\frac{2 M a^{2}}{R^{3}}\right) v^{2}\right] \mathrm{d} t^{2} . \end{align*}dτ2=[(12MR)+4MaR2v(1+a2R2+2Ma2R3)v2]dt2.
Since everything in the square braces is time independent, we have
\begin{align*} \mathrm{d} \tau= & T \end{aligned} \begin{aligned} & \left(1-\frac{2 M}{R}\right)+\frac{4 M a}{R^{2}} \cdot v \\ & \left.-\left(1+\frac{a^{2}}{R^{2}}+\frac{2 M a^{2}}{R^{3}}\right) v^{2}\right]^{\frac{1}{2}} \tag{E.252} \end{align*}\begin{align*} ended with \end{aligned}
with T = 2 π R / v T = 2 π R / v T=2pi R//vT=2 \pi R / vT=2πR/v.
(30.1) Use the equation d t / d s = κ p d t / d s = κ p dt//ds=kappa p\mathrm{d} \boldsymbol{t} / \mathrm{d} s=\kappa \boldsymbol{p}dt/ds=κp and, for a time t t ttt, note that d t / d s = ( d t / d t ) ( d t / d s ) d t / d s = ( d t / d t ) ( d t / d s ) dt//ds=(dt//dt)(dt//ds)\mathrm{d} t / \mathrm{d} s=(\mathrm{d} t / \mathrm{d} t)(\mathrm{d} t / \mathrm{d} s)dt/ds=(dt/dt)(dt/ds), with d s / d t = 1 d s / d t = 1 ds//dt=1\mathrm{d} s / \mathrm{d} t=1ds/dt=1 for motion at unit speed. We therefore have
(E.253) d t d t = κ p (E.253) d t d t = κ p {:(E.253)(dt)/((d)t)=kappa p:}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} t}=\kappa p \tag{E.253} \end{equation*}(E.253)dt dt=κp
which is Newton's second law for a unit mass with velocity t t t\boldsymbol{t}t, subject to a force F = κ p F = κ p F=kappa p\boldsymbol{F}=\kappa \boldsymbol{p}F=κp, where p p p\boldsymbol{p}p is a unit vector perpendicular to the velocity t t ttt and hence also to the curve.
(30.2) Consider the path from pole to equator and back shown in Fig. 11.2, where the vector ends up 90 90 90^(@)90^{\circ}90 out from where it started. We have
(E.254) ( curvature ) = π / 2 ( 1 / 8 ) 4 π a 2 = 1 a 2 (E.254) (  curvature  ) = π / 2 ( 1 / 8 ) 4 π a 2 = 1 a 2 {:(E.254)(" curvature ")=(pi//2)/((1//8)4pia^(2))=(1)/(a^(2)):}\begin{equation*} (\text { curvature })=\frac{\pi / 2}{(1 / 8) 4 \pi a^{2}}=\frac{1}{a^{2}} \tag{E.254} \end{equation*}(E.254)( curvature )=π/2(1/8)4πa2=1a2
(30.3) Start at the origin ( r , θ ) = ( 0 , 0 ) ( r , θ ) = ( 0 , 0 ) (r,theta)=(0,0)(r, \theta)=(0,0)(r,θ)=(0,0). The distance to point ( ε , 0 ) ( ε , 0 ) (epsi,0)(\varepsilon, 0)(ε,0) is
(E.255) r = 0 ε d s = 0 ε d r = ε (E.255) r = 0 ε d s = 0 ε d r = ε {:(E.255)int_(r=0)^(epsi)ds=int_(0)^(epsi)dr=epsi:}\begin{equation*} \int_{r=0}^{\varepsilon} \mathrm{d} s=\int_{0}^{\varepsilon} \mathrm{d} r=\varepsilon \tag{E.255} \end{equation*}(E.255)r=0εds=0εdr=ε
The circumference of a circle with this radius is
θ = 0 2 π d θ sin ε = 2 π sin ε θ = 0 2 π d θ sin ε = 2 π sin ε int_(theta=0)^(2pi)dtheta sin epsi=2pi sin epsi\int_{\theta=0}^{2 \pi} \mathrm{~d} \theta \sin \varepsilon=2 \pi \sin \varepsilonθ=02π dθsinε=2πsinε
We conclude
(E.257) K = lim ε 0 6 ε 2 ( 1 sin ε ε ) = 1 (E.257) K = lim ε 0 6 ε 2 1 sin ε ε = 1 {:(E.257)K=lim_(epsi rarr0)(6)/(epsi^(2))(1-(sin epsi)/(epsi))=1:}\begin{equation*} K=\lim _{\varepsilon \rightarrow 0} \frac{6}{\varepsilon^{2}}\left(1-\frac{\sin \varepsilon}{\varepsilon}\right)=1 \tag{E.257} \end{equation*}(E.257)K=limε06ε2(1sinεε)=1
As this does not vanish, the space is curved.
(30.5) (a) In the final step, we've used the result that
n ^ d t d s = t μ t ν e μ x ν n ^ = t μ t ν K μ ν n ^ d t d s = t μ t ν e μ x ν n ^ = t μ t ν K μ ν {:[ hat(n)*(dt)/((d)s)=t^(mu)t^(nu)(dele_(mu))/(delx^(nu))* hat(n)],[=t^(mu)t^(nu)K_(mu nu)]:}\begin{aligned} \hat{\boldsymbol{n}} \cdot \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s} & =t^{\mu} t^{\nu} \frac{\partial \boldsymbol{e}_{\mu}}{\partial x^{\nu}} \cdot \hat{\boldsymbol{n}} \\ & =t^{\mu} t^{\nu} K_{\mu \nu} \end{aligned}n^dt ds=tμtνeμxνn^=tμtνKμν
(b) Write
g μ σ t μ n σ x ν d x ν d s = K μ ν t μ t ν (E.258) ( g μ σ n σ x ν + K μ ν ) t μ t ν = 0 g μ σ t μ n σ x ν d x ν d s = K μ ν t μ t ν (E.258) g μ σ n σ x ν + K μ ν t μ t ν = 0 {:[g_(mu sigma)t^(mu)(deln^(sigma))/(delx^(nu))*(dx^(nu))/(ds)=-K_(mu nu)t^(mu)t^(nu)],[(E.258)(g_(mu sigma)(deln^(sigma))/(delx^(nu))+K_(mu nu))t^(mu)t^(nu)=0]:}\begin{align*} g_{\mu \sigma} t^{\mu} \frac{\partial n^{\sigma}}{\partial x^{\nu}} \cdot \frac{\mathrm{d} x^{\nu}}{\mathrm{d} s} & =-K_{\mu \nu} t^{\mu} t^{\nu} \\ \left(g_{\mu \sigma} \frac{\partial n^{\sigma}}{\partial x^{\nu}}+K_{\mu \nu}\right) t^{\mu} t^{\nu} & =0 \tag{E.258} \end{align*}gμσtμnσxνdxνds=Kμνtμtν(E.258)(gμσnσxν+Kμν)tμtν=0
(c) Weingarten's equation allows us to say
n ^ x 1 × n ^ x 2 = K 1 μ e μ × K ν 2 ν e ν n ^ x 1 × n ^ x 2 = K 1 μ e μ × K ν 2 ν e ν (del( hat(n)))/(delx^(1))xx(del( hat(n)))/(delx^(2))=K_(1)^(mu)e_(mu)xxK^(nu)_(2)^(nu)e_(nu)\frac{\partial \hat{\boldsymbol{n}}}{\partial x^{1}} \times \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{2}}=K_{1}^{\mu} \boldsymbol{e}_{\mu} \times{K^{\nu}}_{2}^{\nu} \boldsymbol{e}_{\nu}n^x1×n^x2=K1μeμ×Kν2νeν
= det ( K j i ) ( e 1 × e 2 ) = det K j i e 1 × e 2 =det(K_(j)^(i))(e_(1)xxe_(2))=\operatorname{det}\left(K_{j}^{i}\right)\left(e_{1} \times e_{2}\right)=det(Kji)(e1×e2)
Note then that K i j = g i a K a j K i j = g i a K a j K^(i)_(j)=g^(ia)K_(aj)K^{i}{ }_{j}=g^{i a} K_{a j}Kij=giaKaj and so we obtain
(E.260) n ^ ( n ^ x 1 × n ^ x 2 ) = K g | e 1 × e 2 | (E.260) n ^ n ^ x 1 × n ^ x 2 = K g e 1 × e 2 {:(E.260) hat(n)*((del( hat(n)))/(delx^(1))xx(del( hat(n)))/(delx^(2)))=(K)/(g)|e_(1)xxe_(2)|:}\begin{equation*} \hat{\boldsymbol{n}} \cdot\left(\frac{\partial \hat{\boldsymbol{n}}}{\partial x^{1}} \times \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{2}}\right)=\frac{K}{g}\left|\boldsymbol{e}_{1} \times \boldsymbol{e}_{2}\right| \tag{E.260} \end{equation*}(E.260)n^(n^x1×n^x2)=Kg|e1×e2|
(30.7) (a) A computation yields P μ ν P μ ν P^(mu)_(nu)P^{\mu}{ }_{\nu}Pμν.
(b) Compute X ¯ μ n μ X ¯ μ n μ bar(X)^(mu)n_(mu)\bar{X}^{\mu} n_{\mu}X¯μnμ, to show it vanishes. This actually follows from P μ n μ = 0 P μ n μ = 0 P^(mu)n_(mu)=0P^{\mu} n_{\mu}=0Pμnμ=0.
(31.2) A suitable 1-form is W ~ ( ) = f i d x i ( ) W ~ ( ) = f i d x i ( ) tilde(W)()=f_(i)dx^(i)()\tilde{\boldsymbol{W}}()=f_{i} \boldsymbol{d} x^{i}()W~()=fidxi(). Inserting X X vec(X)\vec{X}X we find
(E.261) W ~ ( X ) = f i X i (E.261) W ~ ( X ) = f i X i {:(E.261) tilde(W)( vec(X))=f_(i)X^(i):}\begin{equation*} \tilde{\boldsymbol{W}}(\vec{X})=f_{i} X^{i} \tag{E.261} \end{equation*}(E.261)W~(X)=fiXi
which is the work done.
(31.3) (a) Expanding, we have
T δ ϵ ζ ( α β γ ) = 1 6 ( T δ ϵ ζ α β γ + T δ ϵ ζ α γ β + T δ ϵ ζ β γ α + T β α γ δ ϵ ζ + T δ ϵ ζ γ α β + T γ β β α ) . T δ ϵ ζ ( α β γ ) = 1 6 T δ ϵ ζ α β γ + T δ ϵ ζ α γ β + T δ ϵ ζ β γ α + T β α γ δ ϵ ζ + T δ ϵ ζ γ α β + T γ β β α . {:[T_(delta epsilon zeta)^((alpha beta gamma))=(1)/(6)(T_(delta epsilon zeta)^(alpha beta gamma)+T_(delta epsilon zeta)^(alpha gamma beta)+T_(delta epsilon zeta)^(beta gamma alpha):}],[{:+T^(beta alpha gamma)_(delta epsilon zeta)+T_(delta epsilon zeta)^(gamma alpha beta)+T^(gamma beta beta alpha)).]:}\begin{aligned} T_{\delta \epsilon \zeta}^{(\alpha \beta \gamma)}= & \frac{1}{6}\left(T_{\delta \epsilon \zeta}^{\alpha \beta \gamma}+T_{\delta \epsilon \zeta}^{\alpha \gamma \beta}+T_{\delta \epsilon \zeta}^{\beta \gamma \alpha}\right. \\ & \left.+T^{\beta \alpha \gamma}{ }_{\delta \epsilon \zeta}+T_{\delta \epsilon \zeta}^{\gamma \alpha \beta}+T^{\gamma \beta \beta \alpha}\right) . \end{aligned}Tδϵζ(αβγ)=16(Tδϵζαβγ+Tδϵζαγβ+Tδϵζβγα+Tβαγδϵζ+Tδϵζγαβ+Tγββα).
(E.262)
(b) We find
T [ δ ϵ ζ ] α β γ = 1 6 ( T δ ϵ ζ α β γ T δ ζ ϵ α β γ + T ϵ ζ δ α β γ (E.263) T ϵ δ ζ α β γ + T ζ δ ϵ α β γ T ζ ϵ δ α β γ ) T [ δ ϵ ζ ] α β γ = 1 6 T δ ϵ ζ α β γ T δ ζ ϵ α β γ + T ϵ ζ δ α β γ (E.263) T ϵ δ ζ α β γ + T ζ δ ϵ α β γ T ζ ϵ δ α β γ {:[T_([delta epsilon zeta])^(alpha beta gamma)=(1)/(6)(T_(delta epsilon zeta)^(alpha beta gamma)-T_(delta zeta epsilon)^(alpha beta gamma)+T_(epsilon zeta delta)^(alpha beta gamma):}],[(E.263){:-T_(epsilon delta zeta)^(alpha beta gamma)+T_(zeta delta epsilon)^(alpha beta gamma)-T_(zeta epsilon delta)^(alpha beta gamma))]:}\begin{align*} T_{[\delta \epsilon \zeta]}^{\alpha \beta \gamma}= & \frac{1}{6}\left(T_{\delta \epsilon \zeta}^{\alpha \beta \gamma}-T_{\delta \zeta \epsilon}^{\alpha \beta \gamma}+T_{\epsilon \zeta \delta}^{\alpha \beta \gamma}\right. \\ & \left.-T_{\epsilon \delta \zeta}^{\alpha \beta \gamma}+T_{\zeta \delta \epsilon}^{\alpha \beta \gamma}-T_{\zeta \epsilon \delta}^{\alpha \beta \gamma}\right) \tag{E.263} \end{align*}T[δϵζ]αβγ=16(TδϵζαβγTδζϵαβγ+Tϵζδαβγ(E.263)Tϵδζαβγ+TζδϵαβγTζϵδαβγ)
(31.4) Write
A μ ν T μ ν = 1 2 ( A μ ν T μ ν + A ν μ T ν μ ) A μ ν T μ ν = 1 2 A μ ν T μ ν + A ν μ T ν μ A^(mu nu)T_(mu nu)=(1)/(2)(A^(mu nu)T_(mu nu)+A^(nu mu)T_(nu mu))A^{\mu \nu} T_{\mu \nu}=\frac{1}{2}\left(A^{\mu \nu} T_{\mu \nu}+A^{\nu \mu} T_{\nu \mu}\right)AμνTμν=12(AμνTμν+AνμTνμ)
For (a), this becomes 1 2 A μ ν ( T μ ν + T ν μ ) 1 2 A μ ν T μ ν + T ν μ (1)/(2)A^(mu nu)(T_(mu nu)+T_(nu mu))\frac{1}{2} A^{\mu \nu}\left(T_{\mu \nu}+T_{\nu \mu}\right)12Aμν(Tμν+Tνμ) or A μ ν T ( μ ν ) A μ ν T ( μ ν ) A^(mu nu)T_((mu nu))A^{\mu \nu} T_{(\mu \nu)}AμνT(μν). In the same way, for (b) we get A μ ν T [ μ ν ] A μ ν T [ μ ν ] A^(mu nu)T_([mu nu])A^{\mu \nu} T_{[\mu \nu]}AμνT[μν].
(31.5) (b) For an antisymmetric tensor, the answer simplifies to
(E.265) F [ μ ν ; λ ] = 1 3 ( F μ ν ; λ + F ν λ ; μ + F λ μ ; ν ) (E.265) F [ μ ν ; λ ] = 1 3 F μ ν ; λ + F ν λ ; μ + F λ μ ; ν {:(E.265)F_([mu nu;lambda])=(1)/(3)(F_(mu nu;lambda)+F_(nu lambda;mu)+F_(lambda mu;nu)):}\begin{equation*} F_{[\mu \nu ; \lambda]}=\frac{1}{3}\left(F_{\mu \nu ; \lambda}+F_{\nu \lambda ; \mu}+F_{\lambda \mu ; \nu}\right) \tag{E.265} \end{equation*}(E.265)F[μν;λ]=13(Fμν;λ+Fνλ;μ+Fλμ;ν)
(31.6) Remove mention of the point ξ μ ξ μ xi^(mu)\xi^{\mu}ξμ on either side. Then interpret the combination d / d τ d / d τ d//dtau\mathrm{d} / \mathrm{d} \taud/dτ as a vector. This can be written as d / d τ = ( d x β / d τ ) ( / x β ) d / d τ = d x β / d τ / x β d//dtau=(dx^(beta)//dtau)(del//delx^(beta))\mathrm{d} / \mathrm{d} \tau=\left(\mathrm{d} x^{\beta} / \mathrm{d} \tau\right)\left(\partial / \partial x^{\beta}\right)d/dτ=(dxβ/dτ)(/xβ), interpreted as a set of components multiplied by a set of basis vectors, e β = / x β : e β = / x β : e_(beta)=del//delx^(beta):\boldsymbol{e}_{\beta}=\partial / \partial x^{\beta}:eβ=/xβ:
(E.266) d d τ = d x μ d τ x μ = d x μ d τ e μ (E.266) d d τ = d x μ d τ x μ = d x μ d τ e μ {:(E.266)(d)/((d)tau)=(dx^(mu))/(dtau)(del)/(delx^(mu))=(dx^(mu))/(dtau)e_(mu):}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\partial}{\partial x^{\mu}}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} e_{\mu} \tag{E.266} \end{equation*}(E.266)d dτ=dxμdτxμ=dxμdτeμ
This allows us to write
2 ξ μ x λ x σ = Γ σ λ ν ξ μ x ν 2 x λ x σ = Γ σ λ ν x ν x λ x σ = Γ σ λ ν x ν (E.267) e σ x λ = Γ σ λ ν e ν 2 ξ μ x λ x σ = Γ σ λ ν ξ μ x ν 2 x λ x σ = Γ σ λ ν x ν x λ x σ = Γ σ λ ν x ν (E.267) e σ x λ = Γ σ λ ν e ν {:[(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma))=Gamma_(sigma lambda)^(nu)(delxi^(mu))/(delx^(nu))],[(del^(2))/(delx^(lambda)delx^(sigma))=Gamma_(sigma lambda)^(nu)(del)/(delx^(nu))],[(del)/(delx^(lambda))(del)/(delx^(sigma))=Gamma_(sigma lambda)^(nu)(del)/(delx^(nu))],[(E.267)(dele_(sigma))/(delx^(lambda))=Gamma_(sigma lambda)^(nu)e_(nu)]:}\begin{align*} \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}} & =\Gamma_{\sigma \lambda}^{\nu} \frac{\partial \xi^{\mu}}{\partial x^{\nu}} \\ \frac{\partial^{2}}{\partial x^{\lambda} \partial x^{\sigma}} & =\Gamma_{\sigma \lambda}^{\nu} \frac{\partial}{\partial x^{\nu}} \\ \frac{\partial}{\partial x^{\lambda}} \frac{\partial}{\partial x^{\sigma}} & =\Gamma_{\sigma \lambda}^{\nu} \frac{\partial}{\partial x^{\nu}} \\ \frac{\partial \boldsymbol{e}_{\sigma}}{\partial x^{\lambda}} & =\Gamma_{\sigma \lambda}^{\nu} \boldsymbol{e}_{\nu} \tag{E.267} \end{align*}2ξμxλxσ=Γσλνξμxν2xλxσ=Γσλνxνxλxσ=Γσλνxν(E.267)eσxλ=Γσλνeν
(31.7) (b) Diagram (ii) represents A μ α β B α A μ α β B α A^(mu)_(alpha beta)B^(alpha)A^{\mu}{ }_{\alpha \beta} B^{\alpha}AμαβBα. Diagram (iii) represents A μ α β [ γ C ρ β σ λ ] A μ α β [ γ C ρ β σ λ ] A^(mu)_(alpha beta[gamma)C^(rho beta)_(sigma lambda])A^{\mu}{ }_{\alpha \beta[\gamma} C^{\rho \beta}{ }_{\sigma \lambda]}Aμαβ[γCρβσλ]
(c) Briefly, (iv) represents μ F μ ν = J ν μ F μ ν = J ν grad_(mu)F^(mu nu)=J^(nu)\nabla_{\mu} F^{\mu \nu}=J^{\nu}μFμν=Jν and (v) is
F μ ν ; λ + F ν λ ; μ + F λ μ ; ν = 0 F μ ν ; λ + F ν λ ; μ + F λ μ ; ν = 0 F_(mu nu;lambda)+F_(nu lambda;mu)+F_(lambda mu;nu)=0F_{\mu \nu ; \lambda}+F_{\nu \lambda ; \mu}+F_{\lambda \mu ; \nu}=0Fμν;λ+Fνλ;μ+Fλμ;ν=0. See Chapter 42 for more detail.
(d) The diagram can be written as ( α β β α ) Z μ = α β β α Z μ = (grad_(alpha)grad_(beta)-grad_(beta)grad_(alpha))Z^(mu)=\left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{\nabla}_{\beta}-\boldsymbol{\nabla}_{\beta} \boldsymbol{\nabla}_{\alpha}\right) Z^{\mu}=(αββα)Zμ= R μ ν α β Z ν R μ ν α β Z ν R^(mu)_(nu alpha beta)Z^(nu)R^{\mu}{ }_{\nu \alpha \beta} Z^{\nu}RμναβZν. This expression is examined in Chapter 35.
(32.2) (a) The first equation can be written as
u α a ; α μ = A μ ( u β a β ) + u μ ( A σ a σ ) u α a ; α μ = A μ u β a β + u μ A σ a σ u^(alpha)a_(;alpha)^(mu)=-A^(mu)(u^(beta)a_(beta))+u^(mu)(A^(sigma)a_(sigma))u^{\alpha} a_{; \alpha}^{\mu}=-A^{\mu}\left(u^{\beta} a_{\beta}\right)+u^{\mu}\left(A^{\sigma} a_{\sigma}\right)uαa;αμ=Aμ(uβaβ)+uμ(Aσaσ)
(b) The acceleration A = 0 A = 0 A=0\boldsymbol{A}=0A=0 along a geodesic, so Fermi transport becomes equivalent to parallel transport.
(c) Consider the dot product a b a b a*b\boldsymbol{a} \cdot \boldsymbol{b}ab,
u ( a b ) = ( u a ) b + a ( u b ) = ( A u ) ( b ~ , a ~ ) ( A u ) ( a ~ , b ~ ) = 0 . (E.269) u ( a b ) = u a b + a u b = ( A u ) ( b ~ , a ~ ) ( A u ) ( a ~ , b ~ ) = 0 .  (E.269)  {:[grad_(u)(a*b)=(grad_(u)a)*b+a*(grad_(u)b)],[=-(A^^u)( tilde(b)"," tilde(a))-(A^^u)( tilde(a)"," tilde(b))],[=0.quad" (E.269) "]:}\begin{aligned} \boldsymbol{\nabla}_{u}(\boldsymbol{a} \cdot \boldsymbol{b}) & =\left(\nabla_{u} a\right) \cdot b+\boldsymbol{a} \cdot\left(\nabla_{u} \boldsymbol{b}\right) \\ & =-(\boldsymbol{A} \wedge \boldsymbol{u})(\tilde{b}, \tilde{a})-(\boldsymbol{A} \wedge \boldsymbol{u})(\tilde{\boldsymbol{a}}, \tilde{\boldsymbol{b}}) \\ & =0 . \quad \text { (E.269) } \end{aligned}u(ab)=(ua)b+a(ub)=(Au)(b~,a~)(Au)(a~,b~)=0. (E.269) 
This implies that all angular relationships between the vectors that are Fermi transported are maintained.
(d) Consider the dot product a u a u a*u\boldsymbol{a} \cdot \boldsymbol{u}au,
u ( a u ) = ( u a ) u + a ( u u ) = ( A u ) ( u ~ , a ~ ) + a A = [ A ( u ~ ) u ( a ~ ) u ( u ~ ) A ( a ~ ) ] + a A u ( a u ) = u a u + a u u = ( A u ) ( u ~ , a ~ ) + a A = [ A ( u ~ ) u ( a ~ ) u ( u ~ ) A ( a ~ ) ] + a A {:[grad_(u)(a*u)=(grad_(u)a)*u+a*(grad_(u)u)],[=-(A^^u)( tilde(u)"," tilde(a))+a*A],[=-[A( tilde(u))u( tilde(a))-u( tilde(u))A( tilde(a))]+a*A]:}\begin{aligned} \nabla_{u}(\boldsymbol{a} \cdot \boldsymbol{u}) & =\left(\nabla_{u} a\right) \cdot \boldsymbol{u}+\boldsymbol{a} \cdot\left(\nabla_{u} u\right) \\ & =-(\boldsymbol{A} \wedge \boldsymbol{u})(\tilde{\boldsymbol{u}}, \tilde{a})+\boldsymbol{a} \cdot \boldsymbol{A} \\ & =-[\boldsymbol{A}(\tilde{\boldsymbol{u}}) \boldsymbol{u}(\tilde{a})-\boldsymbol{u}(\tilde{\boldsymbol{u}}) \boldsymbol{A}(\tilde{a})]+\boldsymbol{a} \cdot \boldsymbol{A} \end{aligned}u(au)=(ua)u+a(uu)=(Au)(u~,a~)+aA=[A(u~)u(a~)u(u~)A(a~)]+aA
= A ( a ~ ) + a A = 0 = A ( a ~ ) + a A = 0 =A( tilde(a))+a*A=0=\boldsymbol{A}(\tilde{\boldsymbol{a}})+\boldsymbol{a} \cdot \boldsymbol{A}=0=A(a~)+aA=0
(E.270)
where we have used the facts that: A ( u ) = 0 A ( u ¯ ) = 0 A( bar(u))=0\boldsymbol{A}(\overline{\boldsymbol{u}})=0A(u)=0 and u ( u ) = 1 u ( u ) = 1 u(u)=-1\boldsymbol{u}(\boldsymbol{u})=-1u(u)=1.
(e) Ordinary parallel transport does not maintain the
angle between vectors, but Fermi transport does by virtue of the term ( A u ) ( A u ) -(A^^u)-(\boldsymbol{A} \wedge \boldsymbol{u})(Au), which projects the vector after parallel transport so that it is perpendicular to tangent and relative angles between a , b a , b a,b\boldsymbol{a}, \boldsymbol{b}a,b and c c c\boldsymbol{c}c are maintained.
(33.1) Write
P 4 P 3 = [ u ( P 0 ) + v ( P 1 ) ] [ u ( P 2 ) + v ( P 0 ) ] = [ v ( P 1 ) v ( P 0 ) ] [ u ( P 2 ) u ( P 0 ) ] ( v β x α u α e β ) P 0 ( u β x α v α e β ) P 0 P 4 P 3 = u P 0 + v P 1 u P 2 + v P 0 = v P 1 v P 0 u P 2 u P 0 v β x α u α e β P 0 u β x α v α e β P 0 {:[P_(4)-P_(3)=[u(P_(0))+v(P_(1))]-[u(P_(2))+v(P_(0))]],[=[v(P_(1))-v(P_(0))]-[u(P_(2))-u(P_(0))]],[~~((delv^(beta))/(delx^(alpha))u^(alpha)e_(beta))_(P_(0))-((delu^(beta))/(delx^(alpha))v^(alpha)e_(beta))_(P_(0))]:}\begin{aligned} \mathcal{P}_{4}-\mathcal{P}_{3} & =\left[\boldsymbol{u}\left(\mathcal{P}_{0}\right)+\boldsymbol{v}\left(\mathcal{P}_{1}\right)\right]-\left[\boldsymbol{u}\left(\mathcal{P}_{2}\right)+\boldsymbol{v}\left(\mathcal{P}_{0}\right)\right] \\ & =\left[\boldsymbol{v}\left(\mathcal{P}_{1}\right)-\boldsymbol{v}\left(\mathcal{P}_{0}\right)\right]-\left[\boldsymbol{u}\left(\mathcal{P}_{2}\right)-\boldsymbol{u}\left(\mathcal{P}_{0}\right)\right] \\ & \approx\left(\frac{\partial v^{\beta}}{\partial x^{\alpha}} u^{\alpha} \boldsymbol{e}_{\boldsymbol{\beta}}\right)_{\mathcal{P}_{0}}-\left(\frac{\partial u^{\beta}}{\partial x^{\alpha}} v^{\alpha} \boldsymbol{e}_{\boldsymbol{\beta}}\right)_{\mathcal{P}_{0}} \end{aligned}P4P3=[u(P0)+v(P1)][u(P2)+v(P0)]=[v(P1)v(P0)][u(P2)u(P0)](vβxαuαeβ)P0(uβxαvαeβ)P0
[ u , v ] P 0 [ u , v ] P 0 ~~[u,v]_(P_(0))\approx[\boldsymbol{u}, \boldsymbol{v}]_{\mathcal{P}_{0}}[u,v]P0
(33.2) (a) We find
( £ u A ) ν μ = u α A ν μ x α A ν α u μ x α + A α μ u α x ν £ u A ν μ = u α A ν μ x α A ν α u μ x α + A α μ u α x ν (£_(u)A)_(nu)^(mu)=u^(alpha)(delA_(nu)^(mu))/(delx^(alpha))-A_(nu)^(alpha)(delu^(mu))/(delx^(alpha))+A_(alpha)^(mu)(delu^(alpha))/(delx^(nu))\left(£_{\boldsymbol{u}} \boldsymbol{A}\right)_{\nu}^{\mu}=u^{\alpha} \frac{\partial A_{\nu}^{\mu}}{\partial x^{\alpha}}-A_{\nu}^{\alpha} \frac{\partial u^{\mu}}{\partial x^{\alpha}}+A_{\alpha}^{\mu} \frac{\partial u^{\alpha}}{\partial x^{\nu}}(£uA)νμ=uαAνμxαAναuμxα+Aαμuαxν
(b) Using the previous result
( £ u δ ν μ ) ν μ = u α ( δ ν μ ) x α δ ν α u μ x α + δ α μ u α x ν = u μ x ν + u μ x ν = 0 . £ u δ ν μ ν μ = u α δ ν μ x α δ ν α u μ x α + δ α μ u α x ν = u μ x ν + u μ x ν = 0 . {:[(£_(u)delta_(nu)^(mu))_(nu)^(mu)=u^(alpha)(del(delta_(nu)^(mu)))/(delx^(alpha))-delta_(nu)^(alpha)(delu^(mu))/(delx^(alpha))+delta_(alpha)^(mu)(delu^(alpha))/(delx^(nu))],[=-(delu^(mu))/(delx^(nu))+(delu^(mu))/(delx^(nu))=0.]:}\begin{aligned} \left(£_{\boldsymbol{u}} \delta_{\nu}^{\mu}\right)_{\nu}^{\mu} & =u^{\alpha} \frac{\partial\left(\delta_{\nu}^{\mu}\right)}{\partial x^{\alpha}}-\delta_{\nu}^{\alpha} \frac{\partial u^{\mu}}{\partial x^{\alpha}}+\delta_{\alpha}^{\mu} \frac{\partial u^{\alpha}}{\partial x^{\nu}} \\ & =-\frac{\partial u^{\mu}}{\partial x^{\nu}}+\frac{\partial u^{\mu}}{\partial x^{\nu}}=0 . \end{aligned}(£uδνμ)νμ=uα(δνμ)xαδναuμxα+δαμuαxν=uμxν+uμxν=0.
As a result the contraction (which is equivalent to the delta function) can go either side of the Lie derivative.
(33.4) If u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are Killing fields vectors then
£ u g = £ v g = 0 £ u g = £ v g = 0 £_(u)g=£_(v)g=0£_{\boldsymbol{u}} \boldsymbol{g}=£_{\boldsymbol{v}} \boldsymbol{g}=0£ug=£vg=0
Since £ u £ v £ v £ u = £ [ u , v ] £ u £ v £ v £ u = £ [ u , v ] £_(u)£_(v)-£_(v)£_(u)=£_([u,v])£_{\boldsymbol{u}} £_{\boldsymbol{v}}-£_{\boldsymbol{v}} £_{\boldsymbol{u}}=£_{[\boldsymbol{u}, \boldsymbol{v}]}£u£v£v£u=£[u,v], then we must have
£ [ u , v ] g = 0 £ [ u , v ] g = 0 £_([u,v])g=0£_{[\boldsymbol{u}, \boldsymbol{v}]} \boldsymbol{g}=0£[u,v]g=0
so the commutator is also a Killing vector.
(E.275)
(34.1) We won't assume the commutators vanish. We have that
u + n v v ( u + n ) = [ u + n , v ] = [ u , v ] + [ n , v ] u + n v v ( u + n ) = [ u + n , v ] = [ u , v ] + [ n , v ] grad_(u+n)v-grad_(v)(u+n)=[u+n,v]=[u,v]+[n,v]\nabla_{u+n} v-\nabla_{v}(u+n)=[u+n, v]=[u, v]+[n, v]u+nvv(u+n)=[u+n,v]=[u,v]+[n,v]
= u v v u + n v v n = u v v u + n v v n =grad_(u)v-grad_(v)u+grad_(n)v-grad_(v)n=\nabla_{u} v-\nabla_{v} u+\nabla_{n} v-\nabla_{v} n=uvvu+nvvn
Expanding the second term on the left and rearranging, we have
u + n v v u v n = u v v u + n v v n u + n v v u v n = u v v u + n v v n grad_(u+n)v-grad_(v)u-grad_(v)n=grad_(u)v-grad_(v)u+grad_(n)v-grad_(v)n\nabla_{u+n} v-\nabla_{v} u-\nabla_{v} n=\nabla_{u} v-\nabla_{v} u+\nabla_{n} v-\nabla_{v} nu+nvvuvn=uvvu+nvvn
u + n v = u v + n v u + n v = u v + n v grad_(u+n)v=grad_(u)v+grad_(n)v\nabla_{u+n} v=\nabla_{u} v+\nabla_{n} vu+nv=uv+nv
This is a result that follows from the first rule for the covariant derivative given in the chapter.
(34.3) (a) We have
δ ln det M = ln det ( M + δ M ) ln det M = ln det ( M + δ M ) det M = ln det M 1 ( M + δ M ) = ln det ( I + M 1 δ M ) ln ( I + Tr M 1 δ M ) (E.278) Tr M 1 δ M . δ ln det M _ = ln det ( M _ + δ M _ ) ln det M _ = ln det ( M _ + δ M _ ) det M = ln det M _ 1 ( M _ + δ M _ ) = ln det I + M _ 1 δ M _ ln I + Tr M _ 1 δ M _ (E.278) Tr M _ 1 δ M _ . {:[delta ln detM_=ln det(M_+deltaM_)-ln detM_],[=ln((det(M_+deltaM_))/(det M))],[=ln detM_^(-1)(M_+deltaM_)],[=ln det(I+M_^(-1)deltaM_)],[~~ln(I+TrM_^(-1)deltaM_)],[(E.278)~~TrM_^(-1)deltaM_.]:}\begin{align*} \delta \ln \operatorname{det} \underline{\boldsymbol{M}} & =\ln \operatorname{det}(\underline{\boldsymbol{M}}+\delta \underline{\boldsymbol{M}})-\ln \operatorname{det} \underline{\boldsymbol{M}} \\ & =\ln \frac{\operatorname{det}(\underline{\boldsymbol{M}}+\delta \underline{\boldsymbol{M}})}{\operatorname{det} \boldsymbol{\boldsymbol { M }}} \\ & =\ln \operatorname{det} \underline{\boldsymbol{M}}^{-1}(\underline{\boldsymbol{M}}+\delta \underline{\boldsymbol{M}}) \\ & =\ln \operatorname{det}\left(I+\underline{\boldsymbol{M}}^{-1} \delta \underline{\boldsymbol{M}}\right) \\ & \approx \ln \left(I+\operatorname{Tr} \underline{\boldsymbol{M}}^{-1} \delta \underline{\boldsymbol{M}}\right) \\ & \approx \operatorname{Tr} \underline{\boldsymbol{M}}^{-1} \delta \underline{\boldsymbol{M}} . \tag{E.278} \end{align*}δlndetM=lndet(M+δM)lndetM=lndet(M+δM)detM=lndetM1(M+δM)=lndet(I+M1δM)ln(I+TrM1δM)(E.278)TrM1δM.
(b) Using the identity
(E.279) Γ λ μ μ = 1 2 x λ ln | g | = 1 | g | x λ | g | (E.279) Γ λ μ μ = 1 2 x λ ln | g | = 1 | g | x λ | g | {:(E.279)Gamma_(lambda mu)^(mu)=(1)/(2)(del)/(delx^(lambda))ln |g|=(1)/(sqrt(|g|))(del)/(delx^(lambda))sqrt(|g|):}\begin{equation*} \Gamma_{\lambda \mu}^{\mu}=\frac{1}{2} \frac{\partial}{\partial x^{\lambda}} \ln |g|=\frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{\lambda}} \sqrt{|g|} \tag{E.279} \end{equation*}(E.279)Γλμμ=12xλln|g|=1|g|xλ|g|
(c) Use
(E.280) v ; μ μ = v μ x μ + Γ μ λ μ v λ (E.280) v ; μ μ = v μ x μ + Γ μ λ μ v λ {:(E.280)v_(;mu)^(mu)=(delv^(mu))/(delx^(mu))+Gamma_(mu lambda)^(mu)v^(lambda):}\begin{equation*} v_{; \mu}^{\mu}=\frac{\partial v^{\mu}}{\partial x^{\mu}}+\Gamma_{\mu \lambda}^{\mu} v^{\lambda} \tag{E.280} \end{equation*}(E.280)v;μμ=vμxμ+Γμλμvλ
(34.4) We have
u α f x α = u σ ~ , v u α ( σ μ v μ ) x α = u σ ~ , v + σ ~ , u v u α v μ σ μ x α + u α σ μ v μ x α = v μ ( u σ ~ ) μ + σ μ ( u v ) μ = v μ ( u σ ~ μ ) + σ μ ( u α v μ x α + Γ μ α γ v γ u α ) u α v μ σ μ x α = v μ ( u σ ~ ) μ + Γ μ α γ v γ u α σ μ v μ ( u σ ~ ) μ = u α v μ σ μ x α Γ λ α μ v μ u α σ λ , u α f x α = u σ ~ , v u α σ μ v μ x α = u σ ~ , v + σ ~ , u v u α v μ σ μ x α + u α σ μ v μ x α = v μ u σ ~ μ + σ μ u v μ = v μ u σ ~ μ + σ μ u α v μ x α + Γ μ α γ v γ u α u α v μ σ μ x α = v μ u σ ~ μ + Γ μ α γ v γ u α σ μ v μ u σ ~ μ = u α v μ σ μ x α Γ λ α μ v μ u α σ λ , {:[u^(alpha)(del f)/(delx^(alpha))=grad_(u)(: tilde(sigma)","v:)],[u^(alpha)(del(sigma_(mu)v^(mu)))/(delx^(alpha))=(:grad_(u)( tilde(sigma)),v:)+(:( tilde(sigma)),grad_(u)v:)],[u^(alpha)v^(mu)(delsigma_(mu))/(delx^(alpha))+u^(alpha)sigma_(mu)(delv^(mu))/(delx^(alpha))=v^(mu)(grad_(u)( tilde(sigma)))_(mu)+sigma_(mu)(grad_(u)v)^(mu)],[=v^(mu)(grad_(u) tilde(sigma)_(mu))],[+sigma_(mu)(u^(alpha)(delv^(mu))/(delx^(alpha))+Gamma^(mu)_(alpha gamma)v^(gamma)u^(alpha))],[u^(alpha)v^(mu)(delsigma_(mu))/(delx^(alpha))=v^(mu)(grad_(u)( tilde(sigma)))_(mu)+Gamma^(mu)_(alpha gamma)v^(gamma)u^(alpha)sigma_(mu)],[v^(mu)(grad_(u)( tilde(sigma)))_(mu)=u^(alpha)v^(mu)(delsigma_(mu))/(delx^(alpha))-Gamma^(lambda)_(alpha mu)v^(mu)u^(alpha)sigma_(lambda)","]:}\begin{aligned} u^{\alpha} \frac{\partial f}{\partial x^{\alpha}} & =\nabla_{\boldsymbol{u}}\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle \\ u^{\alpha} \frac{\partial\left(\sigma_{\mu} v^{\mu}\right)}{\partial x^{\alpha}} & =\left\langle\boldsymbol{\nabla}_{u} \tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\right\rangle+\left\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangle \\ u^{\alpha} v^{\mu} \frac{\partial \sigma_{\mu}}{\partial x^{\alpha}}+u^{\alpha} \sigma_{\mu} \frac{\partial v^{\mu}}{\partial x^{\alpha}} & =v^{\mu}\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \tilde{\boldsymbol{\sigma}}\right)_{\mu}+\sigma_{\mu}\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right)^{\mu} \\ & =v^{\mu}\left(\boldsymbol{\nabla}_{u} \tilde{\boldsymbol{\sigma}}_{\mu}\right) \\ & +\sigma_{\mu}\left(u^{\alpha} \frac{\partial v^{\mu}}{\partial x^{\alpha}}+\Gamma^{\mu}{ }_{\alpha \gamma} v^{\gamma} u^{\alpha}\right) \\ u^{\alpha} v^{\mu} \frac{\partial \sigma_{\mu}}{\partial x^{\alpha}} & =v^{\mu}\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \tilde{\boldsymbol{\sigma}}\right)_{\mu}+\Gamma^{\mu}{ }_{\alpha \gamma} v^{\gamma} u^{\alpha} \sigma_{\mu} \\ v^{\mu}\left(\boldsymbol{\nabla}_{u} \tilde{\boldsymbol{\sigma}}\right)_{\mu} & =u^{\alpha} v^{\mu} \frac{\partial \sigma_{\mu}}{\partial x^{\alpha}}-\Gamma^{\lambda}{ }_{\alpha \mu} v^{\mu} u^{\alpha} \sigma_{\lambda}, \end{aligned}uαfxα=uσ~,vuα(σμvμ)xα=uσ~,v+σ~,uvuαvμσμxα+uασμvμxα=vμ(uσ~)μ+σμ(uv)μ=vμ(uσ~μ)+σμ(uαvμxα+Γμαγvγuα)uαvμσμxα=vμ(uσ~)μ+Γμαγvγuασμvμ(uσ~)μ=uαvμσμxαΓλαμvμuασλ,
(E.281)
where, in the last equation, we've relabelled the indices in the final term.
(34.5) (a) Writing components, we find
( u u ) e 1 = u α u μ ; α ( e μ e 1 ) = u α u μ ; α g μ 1 = u α u 1 ; α = u α ( u 1 x α u σ Γ σ α 1 ) = d x α d λ u 1 x α u α u σ Γ σ α 1 (E.282) = d u 1 d λ u α u σ Γ σ α 1 . u u e 1 = u α u μ ; α e μ e 1 = u α u μ ; α g μ 1 = u α u 1 ; α = u α u 1 x α u σ Γ σ α 1 = d x α d λ u 1 x α u α u σ Γ σ α 1 (E.282) = d u 1 d λ u α u σ Γ σ α 1 . {:[(grad_(u)u)*e_(1)=u^(alpha)u^(mu)_(;alpha)(e_(mu)*e_(1))],[=u^(alpha)u^(mu)_(;alpha)g_(mu1)],[=u^(alpha)u_(1;alpha)],[=u^(alpha)((delu_(1))/(delx^(alpha))-u_(sigma)Gamma^(sigma)_(alpha1))],[=(dx^(alpha))/(dlambda)*(delu_(1))/(delx^(alpha))-u^(alpha)u_(sigma)Gamma^(sigma)_(alpha1)],[(E.282)=(du_(1))/((d)lambda)-u^(alpha)u_(sigma)Gamma^(sigma)_(alpha1).]:}\begin{align*} \left(\boldsymbol{\nabla}_{u} \boldsymbol{u}\right) \cdot \boldsymbol{e}_{1} & =u^{\alpha} u^{\mu}{ }_{; \alpha}\left(\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{1}\right) \\ & =u^{\alpha} u^{\mu}{ }_{; \alpha} g_{\mu 1} \\ & =u^{\alpha} u_{1 ; \alpha} \\ & =u^{\alpha}\left(\frac{\partial u_{1}}{\partial x^{\alpha}}-u_{\sigma} \Gamma^{\sigma}{ }_{\alpha 1}\right) \\ & =\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda} \cdot \frac{\partial u_{1}}{\partial x^{\alpha}}-u^{\alpha} u_{\sigma} \Gamma^{\sigma}{ }_{\alpha 1} \\ & =\frac{\mathrm{d} u_{1}}{\mathrm{~d} \lambda}-u^{\alpha} u_{\sigma} \Gamma^{\sigma}{ }_{\alpha 1} . \tag{E.282} \end{align*}(uu)e1=uαuμ;α(eμe1)=uαuμ;αgμ1=uαu1;α=uα(u1xαuσΓσα1)=dxαdλu1xαuαuσΓσα1(E.282)=du1 dλuαuσΓσα1.
This quantity vanishes if u u u\boldsymbol{u}u is the tangent field to a geodesic.
(b) We write u α u σ Γ σ α 1 = u α u σ Γ σ α 1 u α u σ Γ σ α 1 = u α u σ Γ σ α 1 u^(alpha)u_(sigma)Gamma^(sigma)_(alpha1)=u^(alpha)u^(sigma)Gamma_(sigma alpha1)u^{\alpha} u_{\sigma} \Gamma^{\sigma}{ }_{\alpha 1}=u^{\alpha} u^{\sigma} \Gamma_{\sigma \alpha 1}uαuσΓσα1=uαuσΓσα1, then substitute for the connection coefficient in terms of the metric components and make use of the symmetry of the products. A slick method is to directly symmetrize to say
d u 1 d λ = u σ u α Γ ( σ α ) 1 (E.283) = u σ u α 1 2 g 0 α x 1 = 0 d u 1 d λ = u σ u α Γ ( σ α ) 1 (E.283) = u σ u α 1 2 g 0 α x 1 = 0 {:[(du_(1))/((d)lambda)=u^(sigma)u^(alpha)Gamma_((sigma alpha)1)],[(E.283)=u^(sigma)u^(alpha)(1)/(2)(delg_(0alpha))/(delx^(1))=0]:}\begin{align*} \frac{\mathrm{d} u_{1}}{\mathrm{~d} \lambda} & =u^{\sigma} u^{\alpha} \Gamma_{(\sigma \alpha) 1} \\ & =u^{\sigma} u^{\alpha} \frac{1}{2} \frac{\partial g_{0 \alpha}}{\partial x^{1}}=0 \tag{E.283} \end{align*}du1 dλ=uσuαΓ(σα)1(E.283)=uσuα12g0αx1=0
where, in going between lines, we've used g α β , γ = g α β , γ = g_(alpha beta,gamma)=g_{\alpha \beta, \gamma}=gαβ,γ= Γ α β γ + Γ β α γ Γ α β γ + Γ β α γ Gamma_(alpha beta gamma)+Gamma_(beta alpha gamma)\Gamma_{\alpha \beta \gamma}+\Gamma_{\beta \alpha \gamma}Γαβγ+Γβαγ and the fact that g g ggg is independent of x 1 x 1 x^(1)x^{1}x1. (34.6) (a) Combine the result from Example 34.8 with the general rule for forming inner products from eqn 32.21 .
(34.7) To relate the two parametrizations we write
d d λ = f d d s (E.284) d 2 d λ 2 = f d d s + ( f ) 2 d 2 d s 2 d d λ = f d d s (E.284) d 2 d λ 2 = f d d s + f 2 d 2 d s 2 {:[(d)/((d)lambda)=f^(')(d)/((d)s)],[(E.284)(d^(2))/((d)lambda^(2))=f^('')(d)/((d)s)+(f^('))^(2)(d^(2))/((d)s^(2))]:}\begin{align*} \frac{\mathrm{d}}{\mathrm{~d} \lambda} & =f^{\prime} \frac{\mathrm{d}}{\mathrm{~d} s} \\ \frac{\mathrm{~d}^{2}}{\mathrm{~d} \lambda^{2}} & =f^{\prime \prime} \frac{\mathrm{d}}{\mathrm{~d} s}+\left(f^{\prime}\right)^{2} \frac{\mathrm{~d}^{2}}{\mathrm{~d} s^{2}} \tag{E.284} \end{align*}d dλ=fd ds(E.284) d2 dλ2=fd ds+(f)2 d2 ds2
where f = d f d λ f = d f d λ f^(')=(df)/((d)lambda)f^{\prime}=\frac{\mathrm{d} f}{\mathrm{~d} \lambda}f=df dλ. The geodesic equation becomes
(E.285) d 2 x μ d s 2 + f ( f ) 2 d x μ d s + Γ α β μ d x α d s d x β d s = 0 . (E.285) d 2 x μ d s 2 + f f 2 d x μ d s + Γ α β μ d x α d s d x β d s = 0 . {:(E.285)(d^(2)x^(mu))/(ds^(2))+(f^(''))/((f^('))^(2))*((d)x^(mu))/(ds)+Gamma_(alpha beta)^(mu)(dx^(alpha))/(ds)((d)x^(beta))/(ds)=0.:}\begin{equation*} \frac{\mathrm{d}^{2} x^{\mu}}{\mathrm{d} s^{2}}+\frac{f^{\prime \prime}}{\left(f^{\prime}\right)^{2}} \cdot \frac{\mathrm{~d} x^{\mu}}{\mathrm{d} s}+\Gamma_{\alpha \beta}^{\mu} \frac{\mathrm{d} x^{\alpha}}{\mathrm{d} s} \frac{\mathrm{~d} x^{\beta}}{\mathrm{d} s}=0 . \tag{E.285} \end{equation*}(E.285)d2xμds2+f(f)2 dxμds+Γαβμdxαds dxβds=0.
We see that we recover the geodesic equation if the second term vanishes, which is the case if s s sss and λ λ lambda\lambdaλ are linearly related.
(34.8) If the magnitude of u u u\boldsymbol{u}u is constant, then g ( u , u ) g ( u , u ) g(u,u)\boldsymbol{g}(\boldsymbol{u}, \boldsymbol{u})g(u,u) is constant along the curve, and we must have
0 = u α ( g μ ν u μ u ν ) : α = u α u μ ; α u μ + u α u ν u ν ; α (E.286) = 2 u α u μ ; α u μ , 0 = u α g μ ν u μ u ν : α = u α u μ ; α u μ + u α u ν u ν ; α (E.286) = 2 u α u μ ; α u μ , {:[0=u^(alpha)(g_(mu nu)u^(mu)u^(nu))_(:alpha)],[=u^(alpha)u^(mu)_(;alpha)u_(mu)+u^(alpha)u_(nu)u^(nu)_(;alpha)],[(E.286)=2u^(alpha)u^(mu)_(;alpha)u_(mu)","]:}\begin{align*} 0 & =u^{\alpha}\left(g_{\mu \nu} u^{\mu} u^{\nu}\right)_{: \alpha} \\ & =u^{\alpha} u^{\mu}{ }_{; \alpha} u_{\mu}+u^{\alpha} u_{\nu} u^{\nu}{ }_{; \alpha} \\ & =2 u^{\alpha} u^{\mu}{ }_{; \alpha} u_{\mu}, \tag{E.286} \end{align*}0=uα(gμνuμuν):α=uαuμ;αuμ+uαuνuν;α(E.286)=2uαuμ;αuμ,
where we have reindexed the second term in the sum and used g μ ν ; α = 0 g μ ν ; α = 0 g_(mu nu;alpha)=0g_{\mu \nu ; \alpha}=0gμν;α=0. This implies u u u = u u u = u*grad_(u)u=\boldsymbol{u} \cdot \nabla_{\boldsymbol{u}} \boldsymbol{u}=uuu= u α u μ ; α u μ = 0 u α u μ ; α u μ = 0 u^(alpha)u^(mu)_(;alpha)u_(mu)=0u^{\alpha} u^{\mu}{ }_{; \alpha} u_{\mu}=0uαuμ;αuμ=0, even though we don't have the geodesic condition u α u μ ; α = 0 u α u μ ; α = 0 u^(alpha)u^(mu)_(;alpha)=0u^{\alpha} u^{\mu}{ }_{; \alpha}=0uαuμ;α=0.
(35.3) (a) In comma notation, we have
A , ν μ + A , ν ν = J μ A , ν μ + A , ν ν = J μ -A_(,nu)^(mu)+A_(,nu)^(nu)=J^(mu)-A_{, \nu}^{\mu}+A_{, \nu}^{\nu}=J^{\mu}A,νμ+A,νν=Jμ
(b) Since we can write the second term μ ν A ν μ ν A ν del^(mu)del_(nu)A^(nu)\partial^{\mu} \partial_{\nu} A^{\nu}μνAν or ν μ A ν ν μ A ν del_(nu)del^(mu)A^(nu)\partial_{\nu} \partial^{\mu} A^{\nu}νμAν in flat space we can write
(E.288) A ; ν μ ν + A ; ν ν μ = J μ (E.288) A ; ν μ ν + A ; ν ν μ = J μ {:(E.288)-A_(;nu)^(mu)^(nu)+A_(;nu)^(nu)^(mu)=J^(mu):}\begin{equation*} -A_{; \nu}^{\mu}{ }^{\nu}+A_{; \nu}^{\nu}{ }^{\mu}=J^{\mu} \tag{E.288} \end{equation*}(E.288)A;νμν+A;ννμ=Jμ
or
(E.289) A ; ν μ ν + A ν ; μ ν ; μ = J μ (E.289) A ; ν μ ν + A ν ; μ ν ; μ = J μ {:(E.289)-A_(;nu)^(mu)^(nu)+A_(nu;mu)^(nu;mu)=J^(mu):}\begin{equation*} -A_{; \nu}^{\mu}{ }^{\nu}+A_{\nu ; \mu}^{\nu ; \mu}=J^{\mu} \tag{E.289} \end{equation*}(E.289)A;νμν+Aν;μν;μ=Jμ
These will not yield the same answer in curved space, where a non-zero R R R\boldsymbol{R}R tells us that the covariant derivatives don't commute.
(c) The second equation becomes
A ; ν μ ν + A ; ν ν + R ν μ A ν = J μ A ; ν μ ν + A ; ν ν + R ν μ A ν = J μ -A_(;nu)^(mu)^(nu)+A_(;nu)^(nu)+R_(nu)^(mu)A^(nu)=J^(mu)-A_{; \nu}^{\mu}{ }^{\nu}+A_{; \nu}^{\nu}+R_{\nu}^{\mu} A^{\nu}=J^{\mu}A;νμν+A;νν+RνμAν=Jμ
This gives us two versions, with coupling to curvature appearing in one of them through the Ricci tensor.
(35.4) (a) The geodesic equation says
(E.291) d 2 ζ μ d λ 2 + Γ α β μ d ζ α d λ d ζ β d λ = 0 (E.291) d 2 ζ μ d λ 2 + Γ α β μ d ζ α d λ d ζ β d λ = 0 {:(E.291)(d^(2)zeta^(mu))/(dlambda^(2))+Gamma_(alpha beta)^(mu)(dzeta^(alpha))/(dlambda)((d)zeta^(beta))/(dlambda)=0:}\begin{equation*} \frac{\mathrm{d}^{2} \zeta^{\mu}}{\mathrm{d} \lambda^{2}}+\Gamma_{\alpha \beta}^{\mu} \frac{\mathrm{d} \zeta^{\alpha}}{\mathrm{d} \lambda} \frac{\mathrm{~d} \zeta^{\beta}}{\mathrm{d} \lambda}=0 \tag{E.291} \end{equation*}(E.291)d2ζμdλ2+Γαβμdζαdλ dζβdλ=0
Inserting ζ μ = λ u μ ζ μ = λ u μ zeta^(mu)=lambdau^(mu)\zeta^{\mu}=\lambda u^{\mu}ζμ=λuμ gives the quoted equation. Since Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ at the origin can't be a function of the arbitrary directions u μ u μ u^(mu)u^{\mu}uμ, the connection coefficients must vanish at this point.
(c) Reexpress the covariant derivative as
( D n ν d λ ) μ = u α ( n ν μ x α + Γ α β μ n ν β ) = λ n ν μ + u α Γ α β μ n ν β D n ν d λ μ = u α n ν μ x α + Γ α β μ n ν β = λ n ν μ + u α Γ α β μ n ν β {:[((Dn_(nu))/(dlambda))^(mu)=u^(alpha)((deln_(nu)^(mu))/(delx^(alpha))+Gamma_(alpha beta)^(mu)n_(nu)^(beta))],[=(del)/(del lambda)n_(nu)^(mu)+u^(alpha)Gamma_(alpha beta)^(mu)n_(nu)^(beta)]:}\begin{aligned} \left(\frac{D \boldsymbol{n}_{\nu}}{\mathrm{d} \lambda}\right)^{\mu} & =u^{\alpha}\left(\frac{\partial n_{\nu}^{\mu}}{\partial x^{\alpha}}+\Gamma_{\alpha \beta}^{\mu} n_{\nu}^{\beta}\right) \\ & =\frac{\partial}{\partial \lambda} n_{\nu}^{\mu}+u^{\alpha} \Gamma_{\alpha \beta}^{\mu} n_{\nu}^{\beta} \end{aligned}(Dnνdλ)μ=uα(nνμxα+Γαβμnνβ)=λnνμ+uαΓαβμnνβ
and then substitute n ν μ = λ δ ν μ n ν μ = λ δ ν μ n_(nu)^(mu)=lambdadelta_(nu)^(mu)n_{\nu}^{\mu}=\lambda \delta_{\nu}^{\mu}nνμ=λδνμ and the expression for the derivative of the connection coefficients.
(g) Permuting indices gives three expressions. Take one, subtract the second and add the third. Then use the symmetries of the Riemann tensor.
(35.6) (a) We can read off, using ν e μ = Γ α ν μ e α ν e μ = Γ α ν μ e α grad_(nu)e_(mu)=Gamma^(alpha)_(nu mu)e_(alpha)\nabla_{\nu} \boldsymbol{e}_{\mu}=\Gamma^{\alpha}{ }_{\nu \mu} \boldsymbol{e}_{\alpha}νeμ=Γανμeα, that
Γ X X X = 0 , Γ Y X X = 0 Γ X X X = 1 , Γ X Y Y = 1 Γ X X X = 1 , Γ Y Y X = 0 , Γ Y Y Y Γ Y X Y = 1 Y Y Γ X X X = 0 , Γ Y X X = 0 Γ X X X = 1 , Γ X Y Y = 1 Γ X X X = 1 , Γ Y Y X = 0 , Γ Y Y Y Γ Y X Y = 1 Y Y {:[Gamma^(X)_(XX)=0","quadGamma^(Y)_(XX)=0],[Gamma_(XX)^(X)=1","quadGamma_(XY)^(Y)=1],[Gamma_(XX)^(X)=1","],[Gamma_(YY)^(X)=0","quadGamma_(YY)^(Y)Gamma_(YX)^(Y)=-1],[YY]:}\begin{aligned} & \Gamma^{X}{ }_{X X}=0, \quad \Gamma^{Y}{ }_{X X}=0 \\ & \Gamma_{X X}^{X}=1, \quad \Gamma_{X Y}^{Y}=1 \\ & \Gamma_{X X}^{X}=1, \\ & \Gamma_{Y Y}^{X}=0, \quad \Gamma_{Y Y}^{Y} \Gamma_{Y X}^{Y}=-1 \\ & Y Y \end{aligned}ΓXXX=0,ΓYXX=0ΓXXX=1,ΓXYY=1ΓXXX=1,ΓYYX=0,ΓYYYΓYXY=1YY
(b) We find non-zero components
R X X X Y = 1 , R Y X X Y = 1 , R Y X Y X = 1 , R Y X Y Y = 1 , R X X X Y = 1 , R Y X X Y = 1 , R Y X Y X = 1 , R Y X Y Y = 1 , {:[R^(X)_(XXY)=-1","quadR^(Y)_(XXY)=-1","],[R_(YXY)^(X)=-1","quadR_(YXY)^(Y)=1","]:}\begin{aligned} & R^{X}{ }_{X X Y}=-1, \quad R^{Y}{ }_{X X Y}=-1, \\ & R_{Y X Y}^{X}=-1, \quad R_{Y X Y}^{Y}=1, \end{aligned}RXXXY=1,RYXXY=1,RYXYX=1,RYXYY=1,
with other components vanishing.
(36.1) (a) We have
(E.293) ω i ^ t ^ i = ω t ^ i ^ = ω i i ^ i ^ = ω t ^ i ^ (E.293) ω i ^ t ^ i = ω t ^ i ^ = ω i i ^ i ^ = ω t ^ i ^ {:(E.293)omega_( hat(i))^( hat(t)_(i))=-omega_( hat(t) hat(i))=omega_(i hat(i) hat(i))=omega_( hat(t))^( hat(i)):}\begin{equation*} \boldsymbol{\omega}_{\hat{i}}^{\hat{t}_{i}}=-\boldsymbol{\omega}_{\hat{t} \hat{i}}=\boldsymbol{\omega}_{i \hat{i} \hat{i}}=\boldsymbol{\omega}_{\hat{t}}^{\hat{i}} \tag{E.293} \end{equation*}(E.293)ωi^t^i=ωt^i^=ωii^i^=ωt^i^
and (b)
(E.294) ω j ^ i ^ = ω i ^ j ^ = ω j ^ i ^ = ω i ^ j ^ . (E.294) ω j ^ i ^ = ω i ^ j ^ = ω j ^ i ^ = ω i ^ j ^ . {:(E.294)omega_( hat(j))^( hat(i))=omega_( hat(i) hat(j))=-omega_( hat(j) hat(i))=-omega_( hat(i))^( hat(j)).:}\begin{equation*} \boldsymbol{\omega}_{\hat{j}}^{\hat{i}}=\boldsymbol{\omega}_{\hat{i} \hat{j}}=-\omega_{\hat{j} \hat{i}}=-\omega_{\hat{i}}^{\hat{j}} . \tag{E.294} \end{equation*}(E.294)ωj^i^=ωi^j^=ωj^i^=ωi^j^.
(36.2) See the answer to Exercise 10.2.
(36.3) Identify orthonormal components
(E.295) ω σ ^ = d σ , ω ϕ ^ = r ( σ ) d ϕ (E.295) ω σ ^ = d σ , ω ϕ ^ = r ( σ ) d ϕ {:(E.295)omega^( hat(sigma))=d sigma","quadomega^( hat(phi))=r(sigma)d phi:}\begin{equation*} \boldsymbol{\omega}^{\hat{\sigma}}=\boldsymbol{d} \sigma, \quad \boldsymbol{\omega}^{\hat{\phi}}=r(\sigma) \boldsymbol{d} \phi \tag{E.295} \end{equation*}(E.295)ωσ^=dσ,ωϕ^=r(σ)dϕ
and take derivatives
d ω σ ^ = 0 (E.296) d ω ϕ ^ = d r d σ d σ d ϕ = 1 r d r d σ ω ϕ ^ ω σ ^ d ω σ ^ = 0 (E.296) d ω ϕ ^ = d r d σ d σ d ϕ = 1 r d r d σ ω ϕ ^ ω σ ^ {:[domega^( hat(sigma))=0],[(E.296)domega^( hat(phi))=(dr)/((d)sigma)d sigma^^d phi=-(1)/(r)((d)r)/((d)sigma)omega^( hat(phi))^^omega^( hat(sigma))]:}\begin{align*} & \boldsymbol{d} \boldsymbol{\omega}^{\hat{\sigma}}=0 \\ & \boldsymbol{d} \boldsymbol{\omega}^{\hat{\phi}}=\frac{\mathrm{d} r}{\mathrm{~d} \sigma} \boldsymbol{d} \sigma \wedge \boldsymbol{d} \phi=-\frac{1}{r} \frac{\mathrm{~d} r}{\mathrm{~d} \sigma} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\sigma}} \tag{E.296} \end{align*}dωσ^=0(E.296)dωϕ^=dr dσdσdϕ=1r dr dσωϕ^ωσ^
From these, we extract
(E.297) ω σ ^ ϕ ^ = 1 r d r d σ ω ϕ ^ = ω ϕ ^ σ ^ . (E.297) ω σ ^ ϕ ^ = 1 r d r d σ ω ϕ ^ = ω ϕ ^ σ ^ . {:(E.297)omega_( hat(sigma))^( hat(phi))=(1)/(r)((d)r)/((d)sigma)*omega^( hat(phi))=-omega_( hat(phi))^( hat(sigma)).:}\begin{equation*} \boldsymbol{\omega}_{\hat{\sigma}}^{\hat{\phi}}=\frac{1}{r} \frac{\mathrm{~d} r}{\mathrm{~d} \sigma} \cdot \boldsymbol{\omega}^{\hat{\phi}}=-\boldsymbol{\omega}_{\hat{\phi}}^{\hat{\sigma}} . \tag{E.297} \end{equation*}(E.297)ωσ^ϕ^=1r dr dσωϕ^=ωϕ^σ^.
The only thing to do is to take the exterior derivative of this
d ω ϕ ^ ϕ ^ = d ( d r d σ d ϕ ) = d 2 r d σ 2 d σ d ϕ (E.298) = 1 r d 2 r d σ 2 ω σ ^ ω ϕ ^ . d ω ϕ ^ ϕ ^ = d d r d σ d ϕ = d 2 r d σ 2 d σ d ϕ (E.298) = 1 r d 2 r d σ 2 ω σ ^ ω ϕ ^ . {:[domega_( hat(phi))^( hat(phi))=d(-(dr)/((d)sigma)d phi)],[=-(d^(2)r)/((d)sigma^(2))d sigma^^d phi],[(E.298)=-(1)/(r)(d^(2)r)/((d)sigma^(2))omega^( hat(sigma))^^omega^( hat(phi)).]:}\begin{align*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{\phi}}^{\hat{\phi}} & =\boldsymbol{d}\left(-\frac{\mathrm{d} r}{\mathrm{~d} \sigma} \boldsymbol{d} \phi\right) \\ & =-\frac{\mathrm{d}^{2} r}{\mathrm{~d} \sigma^{2}} \boldsymbol{d} \sigma \wedge \boldsymbol{d} \phi \\ & =-\frac{1}{r} \frac{\mathrm{~d}^{2} r}{\mathrm{~d} \sigma^{2}} \boldsymbol{\omega}^{\hat{\sigma}} \wedge \boldsymbol{\omega}^{\hat{\phi}} . \tag{E.298} \end{align*}dωϕ^ϕ^=d(dr dσdϕ)=d2r dσ2dσdϕ(E.298)=1r d2r dσ2ωσ^ωϕ^.
So we have
(E.299) R σ ^ ϕ ^ = 1 r d 2 r d σ 2 ω σ ^ ω ϕ ^ . (E.299) R σ ^ ϕ ^ = 1 r d 2 r d σ 2 ω σ ^ ω ϕ ^ . {:(E.299)R^( hat(sigma))_( hat(phi))=-(1)/(r)(d^(2)r)/((d)sigma^(2))omega^( hat(sigma))^^omega^( hat(phi)).:}\begin{equation*} \mathcal{R}^{\hat{\sigma}}{ }_{\hat{\phi}}=-\frac{1}{r} \frac{\mathrm{~d}^{2} r}{\mathrm{~d} \sigma^{2}} \boldsymbol{\omega}^{\hat{\sigma}} \wedge \boldsymbol{\omega}^{\hat{\phi}} . \tag{E.299} \end{equation*}(E.299)Rσ^ϕ^=1r d2r dσ2ωσ^ωϕ^.
(36.4) We identify orthonormal components
(E.300) ω r ^ = ( 1 + a 2 r 2 ) 1 2 d r , ω θ ^ = r d θ (E.300) ω r ^ = 1 + a 2 r 2 1 2 d r , ω θ ^ = r d θ {:(E.300)omega^( hat(r))=(1+a^(2)r^(2))^((1)/(2))dr","quadomega^( hat(theta))=rd theta:}\begin{equation*} \boldsymbol{\omega}^{\hat{r}}=\left(1+a^{2} r^{2}\right)^{\frac{1}{2}} \boldsymbol{d} r, \quad \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \theta \tag{E.300} \end{equation*}(E.300)ωr^=(1+a2r2)12dr,ωθ^=rdθ
We find connection 1-forms
(E.301) d ω r ^ = 0 , d ω θ ^ = d r d θ = ω θ ^ ω r ^ r ( 1 + a 2 r 2 ) 1 2 (E.301) d ω r ^ = 0 , d ω θ ^ = d r d θ = ω θ ^ ω r ^ r 1 + a 2 r 2 1 2 {:(E.301)domega^( hat(r))=0","quad domega^( hat(theta))=dr^^d theta=-(omega^( hat(theta))^^omega^( hat(r)))/(r(1+a^(2)r^(2))^((1)/(2))):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{r}}=0, \quad \boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}=\boldsymbol{d} r \wedge \boldsymbol{d} \theta=-\frac{\boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{r}}}{r\left(1+a^{2} r^{2}\right)^{\frac{1}{2}}} \tag{E.301} \end{equation*}(E.301)dωr^=0,dωθ^=drdθ=ωθ^ωr^r(1+a2r2)12
We read off that
(E.302) ω r ^ θ ^ = ω θ ^ r ( 1 + a 2 r 2 ) 1 2 = d θ ( 1 + a 2 r 2 ) 1 2 (E.302) ω r ^ θ ^ = ω θ ^ r 1 + a 2 r 2 1 2 = d θ 1 + a 2 r 2 1 2 {:(E.302)omega_( hat(r))^( hat(theta))=(omega^( hat(theta)))/(r(1+a^(2)r^(2))^((1)/(2)))=(d theta)/((1+a^(2)r^(2))^((1)/(2))):}\begin{equation*} \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}}=\frac{\boldsymbol{\omega}^{\hat{\theta}}}{r\left(1+a^{2} r^{2}\right)^{\frac{1}{2}}}=\frac{\boldsymbol{d} \theta}{\left(1+a^{2} r^{2}\right)^{\frac{1}{2}}} \tag{E.302} \end{equation*}(E.302)ωr^θ^=ωθ^r(1+a2r2)12=dθ(1+a2r2)12
so we can calculate
d ω r ^ θ ^ = a 2 r ( 1 + a 2 r 2 ) 3 2 d r d θ (E.303) = a 2 ( 1 + a 2 r 2 ) 2 ω θ ^ ω r ^ d ω r ^ θ ^ = a 2 r 1 + a 2 r 2 3 2 d r d θ (E.303) = a 2 1 + a 2 r 2 2 ω θ ^ ω r ^ {:[domega_( hat(r))^( hat(theta))=-(a^(2)r)/((1+a^(2)r^(2))^((3)/(2)))dr^^d theta],[(E.303)=(a^(2))/((1+a^(2)r^(2))^(2))omega^( hat(theta))^^omega^( hat(r))]:}\begin{align*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}} & =-\frac{a^{2} r}{\left(1+a^{2} r^{2}\right)^{\frac{3}{2}}} \boldsymbol{d} r \wedge \boldsymbol{d} \theta \\ & =\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{r}} \tag{E.303} \end{align*}dωr^θ^=a2r(1+a2r2)32drdθ(E.303)=a2(1+a2r2)2ωθ^ωr^
We have that
(E.304) R r ^ θ ^ = a 2 ( 1 + a 2 r 2 ) 2 ω θ ^ ω r ^ (E.304) R r ^ θ ^ = a 2 1 + a 2 r 2 2 ω θ ^ ω r ^ {:(E.304)R_( hat(r))^( hat(theta))=(a^(2))/((1+a^(2)r^(2))^(2))omega^( hat(theta))^^omega^( hat(r)):}\begin{equation*} \mathcal{R}_{\hat{r}}^{\hat{\theta}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{r}} \tag{E.304} \end{equation*}(E.304)Rr^θ^=a2(1+a2r2)2ωθ^ωr^
so we read off the non-zero component
(E.305) R r ^ | θ ^ r ^ | θ ^ = R r ^ θ ^ r ^ θ ^ = a 2 ( 1 + a 2 r 2 ) 2 , (E.305) R r ^ | θ ^ r ^ | θ ^ = R r ^ θ ^ r ^ θ ^ = a 2 1 + a 2 r 2 2 , {:(E.305)R_( hat(r)| hat(theta) hat(r)|)^( hat(theta))=R_( hat(r) hat(theta) hat(r))^( hat(theta))=(a^(2))/((1+a^(2)r^(2))^(2))",":}\begin{equation*} R_{\hat{r}|\hat{\theta} \hat{r}|}^{\hat{\theta}}=R_{\hat{r} \hat{\theta} \hat{r}}^{\hat{\theta}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}}, \tag{E.305} \end{equation*}(E.305)Rr^|θ^r^|θ^=Rr^θ^r^θ^=a2(1+a2r2)2,
where the | θ ^ r ^ | | θ ^ r ^ | | hat(theta) hat(r)||\hat{\theta} \hat{r}||θ^r^| notation fixes the order of these two variables based on the order in the wedge product in eqn E.304. Note also that since ω θ ^ r ^ = ω r ^ ^ ω θ ^ r ^ = ω r ^ ^ omega_( hat(theta))^( hat(r))=-omega_( hat(hat(r)))\boldsymbol{\omega}_{\hat{\theta}}^{\hat{r}}=-\boldsymbol{\omega}_{\hat{\hat{r}}}ωθ^r^=ωr^^, so we also have
(E.306) R θ ^ r ^ = a 2 ( 1 + a 2 r 2 ) 2 ω r ^ ω θ ^ (E.306) R θ ^ r ^ = a 2 1 + a 2 r 2 2 ω r ^ ω θ ^ {:(E.306)R_( hat(theta))^( hat(r))=(a^(2))/((1+a^(2)r^(2))^(2))omega^( hat(r))^^omega^( hat(theta)):}\begin{equation*} \mathcal{R}_{\hat{\theta}}^{\hat{r}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}} \boldsymbol{\omega}^{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{E.306} \end{equation*}(E.306)Rθ^r^=a2(1+a2r2)2ωr^ωθ^
and
(E.307) R θ ^ | r ^ θ ^ | r ^ = R θ ^ r ^ θ ^ r ^ = a 2 ( 1 + a 2 r 2 ) 2 . (E.307) R θ ^ | r ^ θ ^ | r ^ = R θ ^ r ^ θ ^ r ^ = a 2 1 + a 2 r 2 2 . {:(E.307)R_( hat(theta)| hat(r) hat(theta)|)^( hat(r))=R_( hat(theta) hat(r) hat(theta))^( hat(r))=(a^(2))/((1+a^(2)r^(2))^(2)).:}\begin{equation*} R_{\hat{\theta}|\hat{r} \hat{\theta}|}^{\hat{r}}=R_{\hat{\theta} \hat{r} \hat{\theta}}^{\hat{r}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}} . \tag{E.307} \end{equation*}(E.307)Rθ^|r^θ^|r^=Rθ^r^θ^r^=a2(1+a2r2)2.
This gives a Ricci tensor with components
R r ^ r ^ r ^ = a 2 ( 1 + a 2 r 2 ) 2 , R θ ^ θ ^ = a 2 ( 1 + a 2 r 2 ) 2 . ( E .308 ) R r ^ r ^ r ^ = a 2 1 + a 2 r 2 2 , R θ ^ θ ^ = a 2 1 + a 2 r 2 2 . ( E .308 ) R_( hat(r) hat(r) hat(r))=(a^(2))/((1+a^(2)r^(2))^(2)),quadR_( hat(theta) hat(theta))=(a^(2))/((1+a^(2)r^(2))^(2)).quad(E.308)R_{\hat{r} \hat{r} \hat{r}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}}, \quad R_{\hat{\theta} \hat{\theta}}=\frac{a^{2}}{\left(1+a^{2} r^{2}\right)^{2}} . \quad(\mathrm{E} .308)Rr^r^r^=a2(1+a2r2)2,Rθ^θ^=a2(1+a2r2)2.(E.308)
and a Ricci scalar
(E.309) R = η r r R r r + η θ θ R θ θ = 2 a 2 / ( 1 + a 2 r 2 ) 2 (E.309) R = η r r R r r + η θ θ R θ θ = 2 a 2 / 1 + a 2 r 2 2 {:(E.309)R=eta^(rr)R_(rr)+eta^(theta theta)R_(theta theta)=2a^(2)//(1+a^(2)r^(2))^(2):}\begin{equation*} R=\eta^{r r} R_{r r}+\eta^{\theta \theta} R_{\theta \theta}=2 a^{2} /\left(1+a^{2} r^{2}\right)^{2} \tag{E.309} \end{equation*}(E.309)R=ηrrRrr+ηθθRθθ=2a2/(1+a2r2)2
Note that when r = 0 , R = 2 a 2 r = 0 , R = 2 a 2 r=0,R=2a^(2)r=0, R=2 a^{2}r=0,R=2a2.
(36.5) We identify obvious 1 -forms
(E.310) ω u ^ = ( c + a cos v ) d u , ω v ^ = a d v (E.310) ω u ^ = ( c + a cos v ) d u , ω v ^ = a d v {:(E.310)omega^( hat(u))=(c+a cos v)du","quadomega^( hat(v))=adv:}\begin{equation*} \boldsymbol{\omega}^{\hat{u}}=(c+a \cos v) \boldsymbol{d} u, \quad \boldsymbol{\omega}^{\hat{v}}=a \boldsymbol{d} v \tag{E.310} \end{equation*}(E.310)ωu^=(c+acosv)du,ωv^=adv
Taking exterior derivatives we find
d ω u ^ = a sin v d v d u (E.311) d ω u ^ = 0 d ω u ^ = a sin v d v d u (E.311) d ω u ^ = 0 {:[domega^( hat(u))=-a sin vdv^^du],[(E.311)domega^( hat(u))=0]:}\begin{align*} & \boldsymbol{d} \boldsymbol{\omega}^{\hat{u}}=-a \sin v \boldsymbol{d} v \wedge \boldsymbol{d} u \\ & \boldsymbol{d} \boldsymbol{\omega}^{\hat{u}}=0 \tag{E.311} \end{align*}dωu^=asinvdvdu(E.311)dωu^=0
We conclude that the non-zero curvature 1-form is
(E.312) ω v ^ u ^ = sin v c + a cos v ω u ¯ (E.312) ω v ^ u ^ = sin v c + a cos v ω u ¯ {:(E.312)omega_( hat(v))^( hat(u))=-(sin v)/(c+a cos v)omega^( bar(u)):}\begin{equation*} \boldsymbol{\omega}_{\hat{v}}^{\hat{u}}=-\frac{\sin v}{c+a \cos v} \boldsymbol{\omega}^{\bar{u}} \tag{E.312} \end{equation*}(E.312)ωv^u^=sinvc+acosvωu¯
Now take a second exterior derivative to find
(E.313) d ω v ^ u ^ = cos v a ( c + a cos v ) ω v ^ ω u ^ . (E.313) d ω v ^ u ^ = cos v a ( c + a cos v ) ω v ^ ω u ^ . {:(E.313)domega_( hat(v))^( hat(u))=-(cos v)/(a(c+a cos v))omega^( hat(v))^^omega^( hat(u)).:}\begin{equation*} d \boldsymbol{\omega}_{\hat{v}}^{\hat{u}}=-\frac{\cos v}{a(c+a \cos v)} \boldsymbol{\omega}^{\hat{v}} \wedge \boldsymbol{\omega}^{\hat{u}} . \tag{E.313} \end{equation*}(E.313)dωv^u^=cosva(c+acosv)ωv^ωu^.
As a result we can construct the curvature 2-form, which is
R v ^ u ^ = cos v a ( c + a cos v ) ω u ^ ω v ^ R v ^ u ^ = cos v a ( c + a cos v ) ω u ^ ω v ^ R_( hat(v))^( hat(u))=(cos v)/(a(c+a cos v))omega^( hat(u))^^omega^( hat(v))\mathcal{R}_{\hat{v}}^{\hat{u}}=\frac{\cos v}{a(c+a \cos v)} \boldsymbol{\omega}^{\hat{u}} \wedge \boldsymbol{\omega}^{\hat{v}}Rv^u^=cosva(c+acosv)ωu^ωv^
Assuming an ordering of components ( x 1 , x 2 ) = ( u , v ) x 1 , x 2 = ( u , v ) (x^(1),x^(2))=(u,v)\left(x^{1}, x^{2}\right)=(u, v)(x1,x2)=(u,v) we then extract the components of the Riemann tensor
(E.315) R v ^ u ^ v ^ u ^ = R u ^ v ^ u ^ u ^ = cos v a ( c + a cos v ) . (E.315) R v ^ u ^ v ^ u ^ = R u ^ v ^ u ^ u ^ = cos v a ( c + a cos v ) . {:(E.315)R_( hat(v) hat(u) hat(v))^( hat(u))=R_( hat(u) hat(v) hat(u))^( hat(u))=(cos v)/(a(c+a cos v)).:}\begin{equation*} R_{\hat{v} \hat{u} \hat{v}}^{\hat{u}}=R_{\hat{u} \hat{v} \hat{u}}^{\hat{u}}=\frac{\cos v}{a(c+a \cos v)} . \tag{E.315} \end{equation*}(E.315)Rv^u^v^u^=Ru^v^u^u^=cosva(c+acosv).
This gives us a Ricci tensor with components
(E.316) R u ^ u ^ = R v ^ v ^ = cos v a ( c + a cos v ) (E.316) R u ^ u ^ = R v ^ v ^ = cos v a ( c + a cos v ) {:(E.316)R_( hat(u) hat(u))=R_( hat(v) hat(v))=(cos v)/(a(c+a cos v)):}\begin{equation*} R_{\hat{u} \hat{u}}=R_{\hat{v} \hat{v}}=\frac{\cos v}{a(c+a \cos v)} \tag{E.316} \end{equation*}(E.316)Ru^u^=Rv^v^=cosva(c+acosv)
and consequently a Ricci scalar R = 2 cos v a ( c + a cos v ) R = 2 cos v a ( c + a cos v ) R=(2cos v)/(a(c+a cos v))R=\frac{2 \cos v}{a(c+a \cos v)}R=2cosva(c+acosv).
(36.6) We make the obvious identification
(E.317) ω x ^ = d x r , ω r ^ = d r r (E.317) ω x ^ = d x r , ω r ^ = d r r {:(E.317)omega^( hat(x))=(dx)/(r)","quadomega^( hat(r))=(dr)/(r):}\begin{equation*} \boldsymbol{\omega}^{\hat{x}}=\frac{d x}{r}, \quad \boldsymbol{\omega}^{\hat{r}}=\frac{d r}{r} \tag{E.317} \end{equation*}(E.317)ωx^=dxr,ωr^=drr
The second exterior derivative is
(E.318) d ω r ^ = 0 (E.318) d ω r ^ = 0 {:(E.318)domega^( hat(r))=0:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{r}}=0 \tag{E.318} \end{equation*}(E.318)dωr^=0
so we guess that ω x ^ ^ x ^ ^ ω x ^ ω x ^ ^ x ^ ^ ω x ^ omega_( hat(hat(x)))^( hat(hat(x)))propomega^( hat(x))\boldsymbol{\omega}_{\hat{\hat{x}}}^{\hat{\hat{x}}} \propto \boldsymbol{\omega}^{\hat{x}}ωx^^x^^ωx^. The exterior derivative of the first basis 1 -form is
(E.319) d ω x ^ = 1 r 2 d r d x = ω x ^ ω r ^ (E.319) d ω x ^ = 1 r 2 d r d x = ω x ^ ω r ^ {:(E.319)domega^( hat(x))=-(1)/(r^(2))dr^^dx=-omega^( hat(x))^^omega^( hat(r)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{x}}=-\frac{1}{r^{2}} \boldsymbol{d} r \wedge \boldsymbol{d} x=-\boldsymbol{\omega}^{\hat{x}} \wedge \boldsymbol{\omega}^{\hat{r}} \tag{E.319} \end{equation*}(E.319)dωx^=1r2drdx=ωx^ωr^
and we identify ω x ^ r ^ = ω x ^ ω x ^ r ^ = ω x ^ omega^( hat(x))_( hat(r))=-omega^( hat(x))\boldsymbol{\omega}^{\hat{x}}{ }_{\hat{r}}=-\boldsymbol{\omega}^{\hat{x}}ωx^r^=ωx^ and ω r ^ ^ x ^ = ω x ^ ω r ^ ^ x ^ = ω x ^ omega^( hat(hat(r)))_( hat(x))=omega^( hat(x))\boldsymbol{\omega}^{\hat{\hat{r}}}{ }_{\hat{x}}=\boldsymbol{\omega}^{\hat{x}}ωr^^x^=ωx^. Taking another exterior derivative we have
(E.320) d ω x ^ ^ r ^ = 1 r 2 d r d x = ω r ^ ω x ^ (E.320) d ω x ^ ^ r ^ = 1 r 2 d r d x = ω r ^ ω x ^ {:(E.320)domega_( hat(hat(x)))^( hat(r))=-(1)/(r^(2))dr^^dx=-omega^( hat(r))^^omega^( hat(x)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{\hat{x}}}^{\hat{r}}=-\frac{1}{r^{2}} \boldsymbol{d} r \wedge \boldsymbol{d} x=-\boldsymbol{\omega}^{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{x}} \tag{E.320} \end{equation*}(E.320)dωx^^r^=1r2drdx=ωr^ωx^
which allows us to write
(E.321) R x ^ x ^ ^ = ω y ^ ^ ω x ^ (E.321) R x ^ x ^ ^ = ω y ^ ^ ω x ^ {:(E.321)R_( hat(x))^( hat(hat(x)))=-omega^( hat(hat(y)))^^omega^( hat(x)):}\begin{equation*} \mathcal{R}_{\hat{x}}^{\hat{\hat{x}}}=-\boldsymbol{\omega}^{\hat{\hat{y}}} \wedge \boldsymbol{\omega}^{\hat{x}} \tag{E.321} \end{equation*}(E.321)Rx^x^^=ωy^^ωx^
We find that R r ^ ^ x ^ r ^ x ^ = 1 R r ^ ^ x ^ r ^ x ^ = 1 R^( hat(hat(r)))_( hat(x) hat(r) hat(x))=-1R^{\hat{\hat{r}}}{ }_{\hat{x} \hat{r} \hat{x}}=-1Rr^^x^r^x^=1 and then that R r ^ ^ r ^ x ^ r ^ = 1 R r ^ ^ r ^ x ^ r ^ = 1 R^( hat(hat(r)))_( hat(r) hat(x) hat(r))=-1R^{\hat{\hat{r}}}{ }_{\hat{r} \hat{x} \hat{r}}=-1Rr^^r^x^r^=1. We conclude that the components of the Ricci tensor are
R x ^ x ^ = 1 , R r ^ r ^ = 1 R x ^ x ^ = 1 , R r ^ r ^ = 1 R_( hat(x) hat(x))=-1,quadR_( hat(r) hat(r))=-1R_{\hat{x} \hat{x}}=-1, \quad R_{\hat{r} \hat{r}}=-1Rx^x^=1,Rr^r^=1
(E.322)
meaning that R = 2 R = 2 R=-2R=-2R=2. Converting back to coordinate space
(E.323) R x x = 1 r 2 R r r = 1 r 2 (E.323) R x x = 1 r 2 R r r = 1 r 2 {:(E.323)R_(xx)=-(1)/(r^(2))quadR_(rr)=-(1)/(r^(2)):}\begin{equation*} R_{x x}=-\frac{1}{r^{2}} \quad R_{r r}=-\frac{1}{r^{2}} \tag{E.323} \end{equation*}(E.323)Rxx=1r2Rrr=1r2
(36.7) (a) We identify
(E.324) ω t ^ = d t , ω χ ^ = a d χ ω θ ^ = a sin χ d θ , ω ϕ ^ = a sin χ sin θ d ϕ (E.324) ω t ^ = d t , ω χ ^ = a d χ ω θ ^ = a sin χ d θ , ω ϕ ^ = a sin χ sin θ d ϕ {:(E.324){:[omega^( hat(t))=dt",",omega^( hat(chi))=ad chi],[omega^( hat(theta))=a sin chi d theta",",omega^( hat(phi))=a sin chi sin theta d phi]:}:}\begin{array}{ll} \boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} t, & \boldsymbol{\omega}^{\hat{\chi}}=a \boldsymbol{d} \chi \\ \boldsymbol{\omega}^{\hat{\theta}}=a \sin \chi \boldsymbol{d} \theta, & \boldsymbol{\omega}^{\hat{\phi}}=a \sin \chi \sin \theta \boldsymbol{d} \phi \tag{E.324} \end{array}(E.324)ωt^=dt,ωχ^=adχωθ^=asinχdθ,ωϕ^=asinχsinθdϕ
Since ω t ^ = d t ω t ^ = d t omega^( hat(t))=dt\boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} tωt^=dt we find d ω t ^ = 0 d ω t ^ = 0 domega^( hat(t))=0\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=0dωt^=0. We guess that ω t ^ k ^ ω t ^ k ^ omega^( hat(t))_( hat(k))prop\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}} \proptoωt^k^ ω k ^ ω k ^ omega^( hat(k))\boldsymbol{\omega}^{\hat{k}}ωk^, where k = θ , ϕ k = θ , ϕ k=theta,phik=\theta, \phik=θ,ϕ or χ χ chi\chiχ. (We also have ω k ^ t ^ k ^ = ω k ^ . t ^ ω k ^ t ^ k ^ = ω k ^ . t ^ omega_( hat(k))^( hat(t)_( hat(k)))=omega^( hat(k)). hat(t)\boldsymbol{\omega}_{\hat{k}}^{\hat{t}_{\hat{k}}}=\boldsymbol{\omega}^{\hat{k}} . \hat{t}ωk^t^k^=ωk^.t^.) Another exterior derivative
(E.325) d ω χ ^ = a ˙ d t d χ = a ˙ a ω χ ^ ω t ˙ (E.325) d ω χ ^ = a ˙ d t d χ = a ˙ a ω χ ^ ω t ˙ {:(E.325)domega^( hat(chi))=a^(˙)dt^^d chi=-((a^(˙)))/(a)omega^( hat(chi))^^omega^(t^(˙)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{\chi}}=\dot{a} \boldsymbol{d} t \wedge \boldsymbol{d} \chi=-\frac{\dot{a}}{a} \boldsymbol{\omega}^{\hat{\chi}} \wedge \boldsymbol{\omega}^{\dot{t}} \tag{E.325} \end{equation*}(E.325)dωχ^=a˙dtdχ=a˙aωχ^ωt˙
gives
(E.326) ω t ˙ x ^ ^ = a ˙ a ω x ^ = a ˙ d χ . (E.326) ω t ˙ x ^ ^ = a ˙ a ω x ^ = a ˙ d χ . {:(E.326)omega_(t^(˙))^( hat(hat(x)))=((a^(˙)))/(a)omega^( hat(x))=a^(˙)d_(chi).:}\begin{equation*} \boldsymbol{\omega}_{\dot{t}}^{\hat{\hat{x}}}=\frac{\dot{a}}{a} \boldsymbol{\omega}^{\hat{x}}=\dot{a} \boldsymbol{d}_{\chi} . \tag{E.326} \end{equation*}(E.326)ωt˙x^^=a˙aωx^=a˙dχ.
This fixes the constant of proportionality, so that we assume ω t ^ k ^ = ( a ˙ / a ) ω k ^ ω t ^ k ^ = ( a ˙ / a ) ω k ^ omega^( hat(t))_( hat(k))=(a^(˙)//a)omega^( hat(k))\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}}=(\dot{a} / a) \boldsymbol{\omega}^{\hat{k}}ωt^k^=(a˙/a)ωk^ The next exterior derivative
d ω θ ^ = a ˙ sin χ d t d θ + a cos χ d χ d θ = a ˙ a ω θ ^ ω t ^ 1 a cos χ sin χ ω θ ^ ω χ ^ , gives (E.328) ω χ ^ θ ^ = 1 a cos χ sin χ ω θ ^ = cos χ d θ d ω θ ^ = a ˙ sin χ d t d θ + a cos χ d χ d θ = a ˙ a ω θ ^ ω t ^ 1 a cos χ sin χ ω θ ^ ω χ ^ ,  gives  (E.328) ω χ ^ θ ^ = 1 a cos χ sin χ ω θ ^ = cos χ d θ {:[qquad{:[domega^( hat(theta))=a^(˙)sin chi dt^^d theta+a cos chi d chi^^d theta],[=-((a^(˙)))/(a)omega^( hat(theta))^^omega^( hat(t))-(1)/(a)*(cos chi)/(sin chi)omega^( hat(theta))^^omega^( hat(chi))","]:}],[" gives "],[],[(E.328)quadomega_( hat(chi))^( hat(theta))=(1)/(a)*(cos chi)/(sin chi)omega^( hat(theta))=cos chi d theta]:}\begin{align*} & \qquad \begin{aligned} \boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}} & =\dot{a} \sin \chi \boldsymbol{d} t \wedge \boldsymbol{d} \theta+a \cos \chi \boldsymbol{d} \chi \wedge \boldsymbol{d} \theta \\ & =-\frac{\dot{a}}{a} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{t}}-\frac{1}{a} \cdot \frac{\cos \chi}{\sin \chi} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\chi}}, \end{aligned} \\ & \text { gives } \\ & \\ & \quad \boldsymbol{\omega}_{\hat{\chi}}^{\hat{\theta}}=\frac{1}{a} \cdot \frac{\cos \chi}{\sin \chi} \boldsymbol{\omega}^{\hat{\theta}}=\cos \chi \boldsymbol{d} \theta \tag{E.328} \end{align*}dωθ^=a˙sinχdtdθ+acosχdχdθ=a˙aωθ^ωt^1acosχsinχωθ^ωχ^, gives (E.328)ωχ^θ^=1acosχsinχωθ^=cosχdθ
A final exterior derivative
d ω ϕ ^ = a ˙ sin χ sin θ d t d ϕ + a cos χ sin θ d χ d ϕ (E.329) + a sin χ cos θ d θ d ϕ , d ω ϕ ^ = a ˙ sin χ sin θ d t d ϕ + a cos χ sin θ d χ d ϕ (E.329) + a sin χ cos θ d θ d ϕ , {:[domega^( hat(phi))=a^(˙)sin chi sin theta dt^^d phi+a cos chi sin theta d chi^^d phi],[(E.329)+a sin chi cos theta d theta^^d phi","]:}\begin{align*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{\phi}}= & \dot{a} \sin \chi \sin \theta \boldsymbol{d} t \wedge \boldsymbol{d} \phi+a \cos \chi \sin \theta \boldsymbol{d} \chi \wedge \boldsymbol{d} \phi \\ & +a \sin \chi \cos \theta \boldsymbol{d} \theta \wedge \boldsymbol{d} \phi, \tag{E.329} \end{align*}dωϕ^=a˙sinχsinθdtdϕ+acosχsinθdχdϕ(E.329)+asinχcosθdθdϕ,
gives
(E.330) ω ϕ ^ χ ^ = 1 a cos χ sin χ ω ϕ ^ = cos χ sin θ d ϕ (E.330) ω ϕ ^ χ ^ = 1 a cos χ sin χ ω ϕ ^ = cos χ sin θ d ϕ {:(E.330)omega^( hat(phi)) hat(chi)^(=)(1)/(a)(cos chi)/(sin chi)omega^( hat(phi))=cos chi sin theta d phi:}\begin{equation*} \boldsymbol{\omega}^{\hat{\phi}} \hat{\chi}^{=} \frac{1}{a} \frac{\cos \chi}{\sin \chi} \boldsymbol{\omega}^{\hat{\phi}}=\cos \chi \sin \theta \boldsymbol{d} \phi \tag{E.330} \end{equation*}(E.330)ωϕ^χ^=1acosχsinχωϕ^=cosχsinθdϕ
and
(E.331) ω θ ^ θ ^ A ^ = 1 a cos θ sin χ sin θ ω ϕ ^ = cos θ d ϕ (E.331) ω θ ^ θ ^ A ^ = 1 a cos θ sin χ sin θ ω ϕ ^ = cos θ d ϕ {:(E.331)omega_( hat(theta)_( hat(theta)))^( hat(A))=(1)/(a)(cos theta)/(sin chi sin theta)omega^( hat(phi))=cos theta d phi:}\begin{equation*} \boldsymbol{\omega}_{\hat{\theta}_{\hat{\theta}}}^{\hat{A}}=\frac{1}{a} \frac{\cos \theta}{\sin \chi \sin \theta} \boldsymbol{\omega}^{\hat{\phi}}=\cos \theta \boldsymbol{d} \phi \tag{E.331} \end{equation*}(E.331)ωθ^θ^A^=1acosθsinχsinθωϕ^=cosθdϕ
We're now in a position to compute the components of the curvature 2 -form. These are
R χ ^ t ^ = d ω χ ^ t ^ + ω θ ^ t ^ θ ^ ω χ ^ θ ^ + ω ϕ ^ t ^ ω χ ^ ϕ ^ (E.332) = a ¨ a ω t ^ ω χ ^ with R k ^ t ^ = a ¨ a ω t ^ ω k ^ following. Continuing, R θ ^ χ ^ = d ω θ ^ θ ^ + ω χ ^ ϕ ^ ω θ ^ ϕ ^ + ω t ^ χ ^ χ ^ ω t ^ θ ^ (E.333) = 1 a 2 ω χ ^ ω θ ^ + 0 + ( a ˙ a ) 2 ω χ ^ ω θ ^ , R χ ^ t ^ = d ω χ ^ t ^ + ω θ ^ t ^ θ ^ ω χ ^ θ ^ + ω ϕ ^ t ^ ω χ ^ ϕ ^ (E.332) = a ¨ a ω t ^ ω χ ^  with  R k ^ t ^ = a ¨ a ω t ^ ω k ^  following. Continuing,  R θ ^ χ ^ = d ω θ ^ θ ^ + ω χ ^ ϕ ^ ω θ ^ ϕ ^ + ω t ^ χ ^ χ ^ ω t ^ θ ^ (E.333) = 1 a 2 ω χ ^ ω θ ^ + 0 + a ˙ a 2 ω χ ^ ω θ ^ , {:[R_( hat(chi))^( hat(t))=domega_( hat(chi))^( hat(t))+omega_( hat(theta))^( hat(t)_( hat(theta)))^^omega_( hat(chi))^( hat(theta))+omega_( hat(phi))^( hat(t))^^omega_( hat(chi))^( hat(phi))],[(E.332)=((a^(¨)))/(a)omega^( hat(t))^^omega^( hat(chi))],[" with "R_( hat(k))^( hat(t))=((a^(¨)))/(a)omega^( hat(t))^^omega^( hat(k))" following. Continuing, "],[R_( hat(theta))^( hat(chi))=domega_( hat(theta))_( hat(theta))+omega^( hat(chi))_( hat(phi))^^omega_( hat(theta))^( hat(phi))+omega_( hat(t))^( hat(chi)_( hat(chi)))^^omega^( hat(t))_( hat(theta))],[(E.333)=(1)/(a^(2))omega^( hat(chi))^^omega^( hat(theta))+0+(((a^(˙)))/(a))^(2)omega^( hat(chi))^^omega^( hat(theta))","]:}\begin{gather*} \mathcal{R}_{\hat{\chi}}^{\hat{t}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{\chi}}^{\hat{t}}+\boldsymbol{\omega}_{\hat{\theta}}^{\hat{t}_{\hat{\theta}}} \wedge \boldsymbol{\omega}_{\hat{\chi}}^{\hat{\theta}}+\boldsymbol{\omega}_{\hat{\phi}}^{\hat{t}} \wedge \boldsymbol{\omega}_{\hat{\chi}}^{\hat{\phi}} \\ =\frac{\ddot{a}}{a} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\chi}} \tag{E.332}\\ \text { with } \mathcal{R}_{\hat{k}}^{\hat{t}}=\frac{\ddot{a}}{a} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{k}} \text { following. Continuing, } \\ \mathcal{R}_{\hat{\theta}}^{\hat{\chi}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{\theta}}{ }_{\hat{\theta}}+\boldsymbol{\omega}^{\hat{\chi}}{ }_{\hat{\phi}} \wedge \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}+\boldsymbol{\omega}_{\hat{t}}^{\hat{\chi}_{\hat{\chi}}} \wedge \boldsymbol{\omega}^{\hat{t}}{ }_{\hat{\theta}} \\ =\frac{1}{a^{2}} \boldsymbol{\omega}^{\hat{\chi}} \wedge \boldsymbol{\omega}^{\hat{\theta}}+0+\left(\frac{\dot{a}}{a}\right)^{2} \boldsymbol{\omega}^{\hat{\chi}} \wedge \boldsymbol{\omega}^{\hat{\theta}}, \tag{E.333} \end{gather*}Rχ^t^=dωχ^t^+ωθ^t^θ^ωχ^θ^+ωϕ^t^ωχ^ϕ^(E.332)=a¨aωt^ωχ^ with Rk^t^=a¨aωt^ωk^ following. Continuing, Rθ^χ^=dωθ^θ^+ωχ^ϕ^ωθ^ϕ^+ωt^χ^χ^ωt^θ^(E.333)=1a2ωχ^ωθ^+0+(a˙a)2ωχ^ωθ^,
and also
R j ^ k ^ = 1 + a ˙ 2 a 2 ω k ^ ω j ^ R j ^ k ^ = 1 + a ˙ 2 a 2 ω k ^ ω j ^ R_( hat(j))^( hat(k))=(1+a^(˙)^(2))/(a^(2))omega^( hat(k))^^omega^( hat(j))\mathcal{R}_{\hat{j}}^{\hat{k}}=\frac{1+\dot{a}^{2}}{a^{2}} \boldsymbol{\omega}^{\hat{k}} \wedge \boldsymbol{\omega}^{\hat{j}}Rj^k^=1+a˙2a2ωk^ωj^
The components of the Riemann tensor are then
(E.335) R i i ^ i ^ t ^ = a ¨ a (E.335) R i i ^ i ^ t ^ = a ¨ a {:(E.335)R_(i hat(i) hat(i))^( hat(t))=((a^(¨)))/(a):}\begin{equation*} R_{i \hat{i} \hat{i}}^{\hat{t}}=\frac{\ddot{a}}{a} \tag{E.335} \end{equation*}(E.335)Rii^i^t^=a¨a
and
(E.336) R j ^ i ^ j ^ i ^ = 1 + a ˙ 2 a 2 (E.336) R j ^ i ^ j ^ i ^ = 1 + a ˙ 2 a 2 {:(E.336)R_( hat(j) hat(i) hat(j))^( hat(i))=(1+a^(˙)^(2))/(a^(2)):}\begin{equation*} R_{\hat{j} \hat{i} \hat{j}}^{\hat{i}}=\frac{1+\dot{a}^{2}}{a^{2}} \tag{E.336} \end{equation*}(E.336)Rj^i^j^i^=1+a˙2a2
where we've suspended the summation convention. Now we work out the Ricci tensor. Using R i ^ t ^ i ^ i = R i ^ i ^ i R i ^ t ^ i ^ i = R i ^ i ^ i R^( hat(i) hat(t) hat(i))i=-R_( hat(i) hat(i))^(i)R^{\hat{i} \hat{t} \hat{i}} \boldsymbol{i}=-R_{\hat{i} \hat{i}}^{i}Ri^t^i^i=Ri^i^i (no summation implied), we find
(E.337) R i ^ t ^ = j = 1 3 R t ^ j ^ t ^ j ^ = 3 a ¨ a , and R i ^ i ^ = R i i ^ t ^ t ^ + j ^ = 1 3 R i ^ j ^ i ^ j ^ = a ¨ a + 2 ( 1 + a ˙ 2 a 2 ) (E.337) R i ^ t ^ = j = 1 3 R t ^ j ^ t ^ j ^ = 3 a ¨ a ,  and  R i ^ i ^ = R i i ^ t ^ t ^ + j ^ = 1 3 R i ^ j ^ i ^ j ^ = a ¨ a + 2 1 + a ˙ 2 a 2 {:[(E.337)qquadR_( hat(i) hat(t))=sum_(j=1)^(3)R_( hat(t) hat(j) hat(t))^( hat(j))=-3((a^(¨)))/(a)","],[" and "],[R_( hat(i) hat(i))=R_(i hat(i) hat(t))^( hat(t))+sum_( hat(j)=1)^(3)R_( hat(i) hat(j) hat(i))^( hat(j))=((a^(¨)))/(a)+2((1+a^(˙)^(2))/(a^(2)))]:}\begin{align*} & \qquad R_{\hat{i} \hat{t}}=\sum_{j=1}^{3} R_{\hat{t} \hat{j} \hat{t}}^{\hat{j}}=-3 \frac{\ddot{a}}{a}, \tag{E.337}\\ & \text { and } \\ & R_{\hat{i} \hat{i}}=R_{i \hat{i} \hat{t}}^{\hat{t}}+\sum_{\hat{j}=1}^{3} R_{\hat{i} \hat{j} \hat{i}}^{\hat{j}}=\frac{\ddot{a}}{a}+2\left(\frac{1+\dot{a}^{2}}{a^{2}}\right) \end{align*}(E.337)Ri^t^=j=13Rt^j^t^j^=3a¨a, and Ri^i^=Rii^t^t^+j^=13Ri^j^i^j^=a¨a+2(1+a˙2a2)
(b) Finally, reinstating the summation convention, we work out the R R RRR-scalar
R = η μ ^ ν ^ R μ ^ ν ^ = 3 a ¨ a + 3 a ¨ a + 6 ( 1 + a ˙ 2 a 2 ) (E.339) = 6 ( a ¨ a + 1 a 2 + a ˙ 2 a 2 ) R = η μ ^ ν ^ R μ ^ ν ^ = 3 a ¨ a + 3 a ¨ a + 6 1 + a ˙ 2 a 2 (E.339) = 6 a ¨ a + 1 a 2 + a ˙ 2 a 2 {:[R=eta^( hat(mu) hat(nu))R_( hat(mu) hat(nu))=3((a^(¨)))/(a)+3((a^(¨)))/(a)+6((1+a^(˙)^(2))/(a^(2)))],[(E.339)=6(((a^(¨)))/(a)+(1)/(a^(2))+(a^(˙)^(2))/(a^(2)))]:}\begin{align*} R & =\eta^{\hat{\mu} \hat{\nu}} R_{\hat{\mu} \hat{\nu}}=3 \frac{\ddot{a}}{a}+3 \frac{\ddot{a}}{a}+6\left(\frac{1+\dot{a}^{2}}{a^{2}}\right) \\ & =6\left(\frac{\ddot{a}}{a}+\frac{1}{a^{2}}+\frac{\dot{a}^{2}}{a^{2}}\right) \tag{E.339} \end{align*}R=ημ^ν^Rμ^ν^=3a¨a+3a¨a+6(1+a˙2a2)(E.339)=6(a¨a+1a2+a˙2a2)
(c) We can work out the Einstein tensor, whose components are
(E.340) G t ^ t ^ = 3 ( 1 a 2 + a ˙ 2 a 2 ) (E.340) G t ^ t ^ = 3 1 a 2 + a ˙ 2 a 2 {:(E.340)G_( hat(t) hat(t))=3((1)/(a^(2))+(a^(˙)^(2))/(a^(2))):}\begin{equation*} G_{\hat{t} \hat{t}}=3\left(\frac{1}{a^{2}}+\frac{\dot{a}^{2}}{a^{2}}\right) \tag{E.340} \end{equation*}(E.340)Gt^t^=3(1a2+a˙2a2)
and
(E.341) G i ^ i ^ = ( 2 a ¨ a + 1 a 2 + a ˙ 2 a 2 . ) (E.341) G i ^ i ^ = 2 a ¨ a + 1 a 2 + a ˙ 2 a 2 . {:(E.341)G_( hat(i) hat(i))=-((2(a^(¨)))/(a)+(1)/(a^(2))+(a^(˙)^(2))/(a^(2)).):}\begin{equation*} G_{\hat{i} \hat{i}}=-\left(\frac{2 \ddot{a}}{a}+\frac{1}{a^{2}}+\frac{\dot{a}^{2}}{a^{2}} .\right) \tag{E.341} \end{equation*}(E.341)Gi^i^=(2a¨a+1a2+a˙2a2.)
(36.8) The usual identification gives us 1-forms:
ω T ¯ = e Φ d T , ω R ^ = e Λ d R ω T ¯ = e Φ d T , ω R ^ = e Λ d R omega^( bar(T))=e^(Phi)dT,quadomega^( hat(R))=e^(Lambda)dR\boldsymbol{\omega}^{\bar{T}}=\mathrm{e}^{\Phi} \boldsymbol{d} T, \quad \boldsymbol{\omega}^{\hat{R}}=\mathrm{e}^{\Lambda} \boldsymbol{d} RωT¯=eΦdT,ωR^=eΛdR
(E.342)
Taking derivatives, we obtain
d ω T ^ = Φ e Φ e Λ d T ω R ^ d ω T ^ = Φ e Φ e Λ d T ω R ^ domega^( hat(T))=-Phi^(')e^(Phi)e^(-Lambda)dT^^omega^( hat(R))\boldsymbol{d} \boldsymbol{\omega}^{\hat{T}}=-\Phi^{\prime} \mathrm{e}^{\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{d} T \wedge \boldsymbol{\omega}^{\hat{R}}dωT^=ΦeΦeΛdTωR^
d ω R ^ = Λ ˙ e Φ e Λ d R ω T ^ d ω R ^ = Λ ˙ e Φ e Λ d R ω T ^ domega^( hat(R))=-Lambda^(˙)e^(-Phi)e^(Lambda)dR^^omega^( hat(T))\boldsymbol{d} \boldsymbol{\omega}^{\hat{R}}=-\dot{\Lambda} \mathrm{e}^{-\Phi} \mathrm{e}^{\Lambda} \boldsymbol{d} R \wedge \boldsymbol{\omega}^{\hat{T}}dωR^=Λ˙eΦeΛdRωT^
d ω θ ^ = r ˙ e Φ d θ ω T ^ r e Λ d θ ω R ^ d ω θ ^ = r ˙ e Φ d θ ω T ^ r e Λ d θ ω R ^ domega^( hat(theta))=-r^(˙)e^(-Phi)d theta^^omega^( hat(T))-r^(')e^(-Lambda)d theta^^omega^( hat(R))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}=-\dot{\boldsymbol{r}} \mathrm{e}^{-\Phi} \boldsymbol{d} \theta \wedge \boldsymbol{\omega}^{\hat{T}}-r^{\prime} \mathrm{e}^{-\Lambda} \boldsymbol{d} \theta \wedge \boldsymbol{\omega}^{\hat{R}}dωθ^=r˙eΦdθωT^reΛdθωR^
d ω ϕ ^ = r ˙ e Φ sin θ d ϕ ω T ^ r sin θ e Λ d ϕ ω R ^ d ω ϕ ^ = r ˙ e Φ sin θ d ϕ ω T ^ r sin θ e Λ d ϕ ω R ^ domega^( hat(phi))=-r^(˙)e^(-Phi)sin theta d phi^^omega^( hat(T))-r^(')sin thetae^(-Lambda)d phi^^omega^( hat(R))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\phi}}=-\dot{\boldsymbol{r}} \mathrm{e}^{-\Phi} \sin \theta \boldsymbol{d} \phi \wedge \boldsymbol{\omega}^{\hat{T}}-r^{\prime} \sin \theta \mathrm{e}^{-\Lambda} \boldsymbol{d} \phi \wedge \boldsymbol{\omega}^{\hat{R}}dωϕ^=r˙eΦsinθdϕωT^rsinθeΛdϕωR^
(E.343) cos θ d ϕ ω θ ^ (E.343) cos θ d ϕ ω θ ^ {:(E.343)-cos theta d phi^^omega^( hat(theta)):}\begin{equation*} -\cos \theta \boldsymbol{d} \phi \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{E.343} \end{equation*}(E.343)cosθdϕωθ^
In a slight departure from the other examples, we combine the first two of these to suggest
ω T ^ R ^ = ω R ^ T ^ = Φ e Φ e Λ d T + Λ ˙ e Λ e Φ d R . (E.344) ω T ^ R ^ = ω R ^ T ^ = Φ e Φ e Λ d T + Λ ˙ e Λ e Φ d R .  (E.344)  omega_( hat(T))^( hat(R))=omega_( hat(R))^( hat(T))=Phi^(')e^(Phi)e^(-Lambda)dT+Lambda^(˙)e^(Lambda)e^(-Phi)dR.quad" (E.344) "\boldsymbol{\omega}_{\hat{T}}^{\hat{R}}=\boldsymbol{\omega}_{\hat{R}}^{\hat{T}}=\Phi^{\prime} \mathrm{e}^{\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{d} T+\dot{\Lambda} \mathrm{e}^{\Lambda} \mathrm{e}^{-\Phi} \boldsymbol{d} R . \quad \text { (E.344) }ωT^R^=ωR^T^=ΦeΦeΛdT+Λ˙eΛeΦdR. (E.344) 
This will turn out to be the right choice. The others are then
ω T ^ ^ θ ^ = r ` r e Φ ω θ ^ , ω θ ^ ^ R ^ = r r e Λ ω θ ^ , ω T ^ ϕ = r r r e Φ ω ϕ ^ , ω R ^ ϕ ^ = r e e Λ ω ϕ ^ , ω θ ^ ϕ ^ = cos θ r sin θ ω ϕ ^ . ω T ^ ^ θ ^ = r ` r e Φ ω θ ^ , ω θ ^ ^ R ^ = r r e Λ ω θ ^ , ω T ^ ϕ = r r r e Φ ω ϕ ^ , ω R ^ ϕ ^ = r e e Λ ω ϕ ^ , ω θ ^ ϕ ^ = cos θ r sin θ ω ϕ ^ {:[omega_( hat(hat(T)))^( hat(theta))=((r^(`)))/(r)e^(-Phi)omega^( hat(theta))","quadomega_( hat(hat(theta)))^( hat(R))=(r^('))/(r)e^(-Lambda)omega^( hat(theta))","],[omega_( hat(T))^(phi)=(r_(r))/(r)e^(-Phi)omega^( hat(phi))","quadomega_( hat(R))^( hat(phi))=(r^('))/(e^('))e^(-Lambda)omega^( hat(phi))","],[omega_( hat(theta))^( hat(phi))=(cos theta)/(r sin theta)omega^( hat(phi))". "]:}\begin{aligned} & \boldsymbol{\omega}_{\hat{\hat{T}}}^{\hat{\theta}}=\frac{\grave{r}}{r} \mathrm{e}^{-\Phi} \boldsymbol{\omega}^{\hat{\theta}}, \quad \boldsymbol{\omega}_{\hat{\hat{\theta}}}^{\hat{R}}=\frac{r^{\prime}}{r} \mathrm{e}^{-\Lambda} \boldsymbol{\omega}^{\hat{\theta}}, \\ & \boldsymbol{\omega}_{\hat{T}}^{\phi}=\frac{r_{r}}{r} \mathrm{e}^{-\Phi} \boldsymbol{\omega}^{\hat{\phi}}, \quad \boldsymbol{\omega}_{\hat{R}}^{\hat{\phi}}=\frac{r^{\prime}}{\mathrm{e}^{\prime}} \mathrm{e}^{-\Lambda} \boldsymbol{\omega}^{\hat{\phi}}, \\ & \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}=\frac{\cos \theta}{r \sin \theta} \boldsymbol{\omega}^{\hat{\phi}} \text {. } \end{aligned}ωT^^θ^=r`reΦωθ^,ωθ^^R^=rreΛωθ^,ωT^ϕ=rrreΦωϕ^,ωR^ϕ^=reeΛωϕ^,ωθ^ϕ^=cosθrsinθωϕ^
The derivatives are
d ω R ^ R ^ = { [ Φ + ( Φ ) 2 Φ Λ ] e 2 Λ + [ Λ ¨ + ( Λ ˙ ) Λ ˙ Φ ˙ e 2 Φ ] } ω θ ^ ω R ^ , d ω R ^ R ^ = Φ + Φ 2 Φ Λ e 2 Λ + Λ ¨ + ( Λ ˙ ) Λ ˙ Φ ˙ e 2 Φ ω θ ^ ω R ^ , {:[domega_( hat(R))^( hat(R))={-[Phi^('')+(Phi^('))^(2)-Phi^(')Lambda^(')]e^(-2Lambda):}],[{:+[(Lambda^(¨))+((Lambda^(˙)))-(Lambda^(˙))(Phi^(˙))e^(-2Phi)]}omega^( hat(theta))^^omega^( hat(R))","]:}\begin{aligned} & d \boldsymbol{\omega}_{\hat{R}}^{\hat{R}}=\left\{-\left[\Phi^{\prime \prime}+\left(\Phi^{\prime}\right)^{2}-\Phi^{\prime} \Lambda^{\prime}\right] \mathrm{e}^{-2 \Lambda}\right. \\ & \left.+\left[\ddot{\Lambda}+(\dot{\Lambda})-\dot{\Lambda} \dot{\Phi} \mathrm{e}^{-2 \Phi}\right]\right\} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{R}}, \end{aligned}dωR^R^={[Φ+(Φ)2ΦΛ]e2Λ+[Λ¨+(Λ˙)Λ˙Φ˙e2Φ]}ωθ^ωR^,
d ω T ^ θ ^ = 1 r ( r ¨ + r ˙ Φ ˙ ) e 2 Φ ω θ ^ ω T ^ d ω T ^ θ ^ = 1 r ( r ¨ + r ˙ Φ ˙ ) e 2 Φ ω θ ^ ω T ^ domega_( hat(T))^( hat(theta))=(1)/(r)(-r^(¨)+r^(˙)Phi^(˙))e^(-2Phi)omega^( hat(theta))^^omega^( hat(T))\boldsymbol{d} \boldsymbol{\omega}_{\hat{T}}^{\hat{\theta}}=\frac{1}{r}(-\ddot{r}+\dot{r} \dot{\Phi}) \mathrm{e}^{-2 \Phi} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{T}}dωT^θ^=1r(r¨+r˙Φ˙)e2Φωθ^ωT^
+ 1 r ( r ˙ + r ˙ Φ ) e Φ e Λ ω θ ^ ω R ^ + 1 r r ˙ + r ˙ Φ e Φ e Λ ω θ ^ ω R ^ +(1)/(r)(-r^(˙)^(')+(r^(˙))Phi^('))e^(-Phi)e^(-Lambda)omega^( hat(theta))^^omega^( hat(R))+\frac{1}{r}\left(-\dot{r}^{\prime}+\dot{r} \Phi^{\prime}\right) \mathrm{e}^{-\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{R}}+1r(r˙+r˙Φ)eΦeΛωθ^ωR^
d ω R ^ θ ^ = 1 r ( r ˙ r Λ ˙ ) e Φ e Λ ω θ ^ ω T ^ d ω R ^ θ ^ = 1 r r ˙ r Λ ˙ e Φ e Λ ω θ ^ ω T ^ domega_( hat(R))^( hat(theta))=-(1)/(r)(r^(˙)^(')-r^(')(Lambda^(˙)))e^(-Phi)e^(-Lambda)omega^( hat(theta))^^omega^( hat(T))\boldsymbol{d} \boldsymbol{\omega}_{\hat{R}}^{\hat{\theta}}=-\frac{1}{r}\left(\dot{r}^{\prime}-r^{\prime} \dot{\Lambda}\right) \mathrm{e}^{-\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{T}}dωR^θ^=1r(r˙rΛ˙)eΦeΛωθ^ωT^
1 r ( r r Λ ) e 2 Λ ω θ ^ ω R ^ 1 r r r Λ e 2 Λ ω θ ^ ω R ^ -(1)/(r)(r^('')-r^(')Lambda^('))e^(-2Lambda)omega^( hat(theta))^^omega^( hat(R))-\frac{1}{r}\left(r^{\prime \prime}-r^{\prime} \Lambda^{\prime}\right) \mathrm{e}^{-2 \Lambda} \omega^{\hat{\theta}} \wedge \omega^{\hat{R}}1r(rrΛ)e2Λωθ^ωR^
d ω T ^ ϕ ^ = 1 r ( r ¨ r ˙ Φ ˙ ) e 2 Φ ω ϕ ^ ω T ^ d ω T ^ ϕ ^ = 1 r ( r ¨ r ˙ Φ ˙ ) e 2 Φ ω ϕ ^ ω T ^ domega_( hat(T))^( hat(phi))=-(1)/(r)(r^(¨)-r^(˙)Phi^(˙))e^(-2Phi)omega^( hat(phi))^^omega^( hat(T))\boldsymbol{d} \boldsymbol{\omega}_{\hat{T}}^{\hat{\phi}}=-\frac{1}{r}(\ddot{r}-\dot{r} \dot{\Phi}) \mathrm{e}^{-2 \Phi} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{T}}dωT^ϕ^=1r(r¨r˙Φ˙)e2Φωϕ^ωT^
1 r ( r ˙ r ˙ Φ ) e Λ e Φ ω ϕ ^ ω R ^ 1 r r ˙ r ˙ Φ e Λ e Φ ω ϕ ^ ω R ^ -(1)/(r)(r^(˙)^(')-(r^(˙))Phi^('))e^(-Lambda)e^(-Phi)omega^( hat(phi))^^omega^( hat(R))-\frac{1}{r}\left(\dot{r}^{\prime}-\dot{r} \Phi^{\prime}\right) \mathrm{e}^{-\Lambda} \mathrm{e}^{-\Phi} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{R}}1r(r˙r˙Φ)eΛeΦωϕ^ωR^
r ˙ cos θ r 2 sin θ e Φ ω ϕ ^ ω θ ^ r ˙ cos θ r 2 sin θ e Φ ω ϕ ^ ω θ ^ -((r^(˙))cos theta)/(r^(2)sin theta)e^(-Phi)omega^( hat(phi))^^omega^( hat(theta))-\frac{\dot{r} \cos \theta}{r^{2} \sin \theta} \mathrm{e}^{-\Phi} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}}r˙cosθr2sinθeΦωϕ^ωθ^
(E.348)
d ω R ^ ϕ ^ = 1 r ( r ˙ r Λ ˙ ) e Λ e Φ ω ϕ ^ ω T ^ d ω R ^ ϕ ^ = 1 r r ˙ r Λ ˙ e Λ e Φ ω ϕ ^ ω T ^ domega_( hat(R))^( hat(phi))=-(1)/(r)(r^(˙)^(')-r^(')(Lambda^(˙)))e^(-Lambda)e^(-Phi)omega^( hat(phi))^^omega^( hat(T))\boldsymbol{d} \boldsymbol{\omega}_{\hat{R}}^{\hat{\phi}}=-\frac{1}{r}\left(\dot{r}^{\prime}-r^{\prime} \dot{\Lambda}\right) \mathrm{e}^{-\Lambda} \mathrm{e}^{-\Phi} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{T}}dωR^ϕ^=1r(r˙rΛ˙)eΛeΦωϕ^ωT^
1 r ( r r Λ ˙ ) e 2 Λ ω ϕ ^ ω R ^ 1 r r r Λ ˙ e 2 Λ ω ϕ ^ ω R ^ -(1)/(r)(r^('')-r^(')(Lambda^(˙)))e^(-2Lambda)omega^( hat(phi))^^omega^( hat(R))-\frac{1}{r}\left(r^{\prime \prime}-r^{\prime} \dot{\Lambda}\right) \mathrm{e}^{-2 \Lambda} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{R}}1r(rrΛ˙)e2Λωϕ^ωR^
(E.349) r cos θ r 2 sin θ e Λ ω ϕ ^ ω θ ^ (E.349) r cos θ r 2 sin θ e Λ ω ϕ ^ ω θ ^ {:(E.349)-(r^(')cos theta)/(r^(2)sin theta)e^(-Lambda)omega^( hat(phi))^^omega^( hat(theta)):}\begin{equation*} -\frac{r^{\prime} \cos \theta}{r^{2} \sin \theta} \mathrm{e}^{-\Lambda} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{E.349} \end{equation*}(E.349)rcosθr2sinθeΛωϕ^ωθ^
and
(E.350) d ω θ ^ ϕ ^ = 1 r 2 ω ϕ ^ ω θ ^ (E.350) d ω θ ^ ϕ ^ = 1 r 2 ω ϕ ^ ω θ ^ {:(E.350)domega_( hat(theta))^( hat(phi))=(1)/(r^(2))omega^( hat(phi))^^omega^( hat(theta)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}=\frac{1}{r^{2}} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{E.350} \end{equation*}(E.350)dωθ^ϕ^=1r2ωϕ^ωθ^
We then have non-zero contributions to R ν ^ μ ^ R ν ^ μ ^ R_( hat(nu))^( hat(mu))\mathcal{R}_{\hat{\nu}}^{\hat{\mu}}Rν^μ^ of
and substitution of these gives the required results.
(37.1) (a) Insert the velocity and we have σ ~ ( u ) = V u α u α = σ ~ ( u ) = V u α u α = tilde(sigma)(u)=-Vu^(alpha)u_(alpha)=\tilde{\boldsymbol{\sigma}}(\boldsymbol{u})=-V u^{\alpha} u_{\alpha}=σ~(u)=Vuαuα=
V V VVV, as required for σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~.
(b) The total momentum in the box is
p = V T ( , u ~ ) = V T μ ν u ν e μ p = V T ( , u ~ ) = V T μ ν u ν e μ p=-VT(, tilde(u))=-VT^(mu nu)u_(nu)e_(mu)\boldsymbol{p}=-V \boldsymbol{T}(, \tilde{\boldsymbol{u}})=-V T^{\mu \nu} u_{\nu} \boldsymbol{e}_{\mu}p=VT(,u~)=VTμνuνeμ
(E.352)
(c) The total energy is
E = p u = V T μ ν u μ ( g σ ν u σ ) (E.353) = V T μ ν u μ u ν = V T ( u ~ , u ~ ) E = p u = V T μ ν u μ g σ ν u σ (E.353) = V T μ ν u μ u ν = V T ( u ~ , u ~ ) {:[E=-p*u=VT^(mu nu)u_(mu)(g_(sigma nu)u^(sigma))],[(E.353)=VT^(mu nu)u_(mu)u_(nu)=VT( tilde(u)"," tilde(u))]:}\begin{align*} E & =-\boldsymbol{p} \cdot \boldsymbol{u}=V T^{\mu \nu} u_{\mu}\left(g_{\sigma \nu} u^{\sigma}\right) \\ & =V T^{\mu \nu} u_{\mu} u_{\nu}=V \boldsymbol{T}(\tilde{\boldsymbol{u}}, \tilde{\boldsymbol{u}}) \tag{E.353} \end{align*}E=pu=VTμνuμ(gσνuσ)(E.353)=VTμνuμuν=VT(u~,u~)
(37.2) (a) Since the box lives in the observer's three space, the tetrad ( u , A , B , C ) u , A , B , C ) u,A,B,C)\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})u,A,B,C) spans spacetime.
(b) Inserting the velocity 1 -form into the current gives n = J ( u ~ ) n = J ( u ~ ) n=-J( tilde(u))n=-\boldsymbol{J}(\tilde{\boldsymbol{u}})n=J(u~). We can then write the current vector as
J = n u + a A + b B + c C J = n u + a A + b B + c C J=nu+aA+bB+cC\boldsymbol{J}=n \boldsymbol{u}+a \boldsymbol{A}+b \boldsymbol{B}+c \boldsymbol{C}J=nu+aA+bB+cC
(E.354)
R T ^ R ^ = d ω T ^ R ^ , R T ^ θ ^ = d ω T ^ θ ^ + ω T ^ R ^ ω R ^ θ ^ , R T ^ ϕ ^ = d ω T ^ ϕ ^ + ω T ^ R ^ ω R ^ ϕ ^ + ω T ^ θ ^ ω θ ^ ϕ ^ , R θ ^ ϕ ^ = d ω θ ^ ϕ ^ + ω θ ^ T ^ ω T ^ ϕ ^ + ω θ ^ R ^ ω R ^ ϕ ^ , R R ^ θ ^ = d ω R ^ θ ^ θ ^ + ω R ^ T ^ ω T ^ θ ^ θ ^ , R R ^ ϕ ^ = d ω R ^ ϕ ^ + ω R ^ T ^ ω T ^ ϕ ^ + ω R ^ θ ^ ω θ ^ ϕ ^ , (E.351) R T ^ R ^ = d ω T ^ R ^ , R T ^ θ ^ = d ω T ^ θ ^ + ω T ^ R ^ ω R ^ θ ^ , R T ^ ϕ ^ = d ω T ^ ϕ ^ + ω T ^ R ^ ω R ^ ϕ ^ + ω T ^ θ ^ ω θ ^ ϕ ^ , R θ ^ ϕ ^ = d ω θ ^ ϕ ^ + ω θ ^ T ^ ω T ^ ϕ ^ + ω θ ^ R ^ ω R ^ ϕ ^ , R R ^ θ ^ = d ω R ^ θ ^ θ ^ + ω R ^ T ^ ω T ^ θ ^ θ ^ , R R ^ ϕ ^ = d ω R ^ ϕ ^ + ω R ^ T ^ ω T ^ ϕ ^ + ω R ^ θ ^ ω θ ^ ϕ ^ ,  (E.351)  {:[R^( hat(T))_( hat(R))=domega^( hat(T))_( hat(R))","],[R^( hat(T))_( hat(theta))=domega^( hat(T))_( hat(theta))+omega^( hat(T))_( hat(R))^^omega^( hat(R))_( hat(theta))","],[R^( hat(T))_( hat(phi))=domega^( hat(T))_( hat(phi))+omega^( hat(T))_( hat(R))^^omega^( hat(R))_( hat(phi))+omega^( hat(T))_( hat(theta))^^omega^( hat(theta))_( hat(phi))","],[R^( hat(theta))_( hat(phi))=domega^( hat(theta))_( hat(phi))+omega^( hat(theta))_( hat(T))^^omega^( hat(T))_( hat(phi))+omega^( hat(theta))_( hat(R))^^omega^( hat(R))_( hat(phi))","],[R^( hat(R))_( hat(theta))=domega^( hat(R)) hat(theta)_( hat(theta))+omega^( hat(R))_( hat(T))^^omega^( hat(T)) hat(theta)_( hat(theta))","],[R^( hat(R))_( hat(phi))=domega^( hat(R))_( hat(phi))+omega^( hat(R))_( hat(T))^^omega^( hat(T))_( hat(phi))+omega^( hat(R))_( hat(theta))^^omega^( hat(theta))_( hat(phi))","quad" (E.351) "]:}\begin{aligned} & \mathcal{R}^{\hat{T}}{ }_{\hat{R}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{T}}{ }_{\hat{R}}, \\ & \mathcal{R}^{\hat{T}}{ }_{\hat{\theta}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{T}}{ }_{\hat{\theta}}+\boldsymbol{\omega}^{\hat{T}}{ }_{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{R}}{ }_{\hat{\theta}}, \\ & \mathcal{R}^{\hat{T}}{ }_{\hat{\phi}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{T}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{T}}{ }_{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{R}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{T}}{ }_{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{\phi}}, \\ & \mathcal{R}^{\hat{\theta}}{ }_{\hat{\phi}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{T}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{R}}{ }_{\hat{\phi}}, \\ & \mathcal{R}^{\hat{R}}{ }_{\hat{\theta}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{R}} \hat{\theta}_{\hat{\theta}}+\boldsymbol{\omega}^{\hat{R}}{ }_{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{T}} \hat{\theta}_{\hat{\theta}}, \\ & \mathcal{R}^{\hat{R}}{ }_{\hat{\phi}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{R}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{R}}{ }_{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{T}}{ }_{\hat{\phi}}+\boldsymbol{\omega}^{\hat{R}}{ }_{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{\phi}}, \quad \text { (E.351) } \end{aligned}RT^R^=dωT^R^,RT^θ^=dωT^θ^+ωT^R^ωR^θ^,RT^ϕ^=dωT^ϕ^+ωT^R^ωR^ϕ^+ωT^θ^ωθ^ϕ^,Rθ^ϕ^=dωθ^ϕ^+ωθ^T^ωT^ϕ^+ωθ^R^ωR^ϕ^,RR^θ^=dωR^θ^θ^+ωR^T^ωT^θ^θ^,RR^ϕ^=dωR^ϕ^+ωR^T^ωT^ϕ^+ωR^θ^ωθ^ϕ^, (E.351) 
The volume of the box is
(E.355) V = ω ~ ( u , A , B , C ) (E.355) V = ω ~ ( u , A , B , C ) {:(E.355)V= tilde(omega)(u","A","B","C):}\begin{equation*} V=\tilde{\boldsymbol{\omega}}(\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}) \tag{E.355} \end{equation*}(E.355)V=ω~(u,A,B,C)
We then have
ω ~ ( J , A , B , C ) = n ω ~ ( u , A , B , C ) (E.356) = n V = N ω ~ ( J , A , B , C ) = n ω ~ ( u , A , B , C ) (E.356) = n V = N {:[ tilde(omega)(J","A","B","C)=n tilde(omega)(u","A","B","C)],[(E.356)=nV=N]:}\begin{align*} \tilde{\boldsymbol{\omega}}(\boldsymbol{J}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}) & =n \tilde{\boldsymbol{\omega}}(\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}) \\ & =n V=N \tag{E.356} \end{align*}ω~(J,A,B,C)=nω~(u,A,B,C)(E.356)=nV=N
(c) If we choose
A = A + α J B = B + β J (E.357) C = C + γ J A = A + α J B = B + β J (E.357) C = C + γ J {:[A^(')=A+alpha J],[B^(')=B+beta J],[(E.357)C^(')=C+gamma J]:}\begin{align*} & \boldsymbol{A}^{\prime}=\boldsymbol{A}+\alpha \boldsymbol{J} \\ & \boldsymbol{B}^{\prime}=\boldsymbol{B}+\beta \boldsymbol{J} \\ & \boldsymbol{C}^{\prime}=\boldsymbol{C}+\gamma \boldsymbol{J} \tag{E.357} \end{align*}A=A+αJB=B+βJ(E.357)C=C+γJ
where α , β α , β alpha,beta\alpha, \betaα,β and γ γ gamma\gammaγ are constants, then each box captures the same number of particles.
(37.5) (a) We have, by definition of the covariant derivative,
(E.358) μ ω ~ = d ω ~ , e μ = 0 (E.358) μ ω ~ = d ω ~ , e μ = 0 {:(E.358)grad_(mu) tilde(omega)=(:d( tilde(omega)),e_(mu):)=0:}\begin{equation*} \nabla_{\mu} \tilde{\boldsymbol{\omega}}=\left\langle\boldsymbol{d} \tilde{\boldsymbol{\omega}}, \boldsymbol{e}_{\mu}\right\rangle=0 \tag{E.358} \end{equation*}(E.358)μω~=dω~,eμ=0
(c) From part (b) we have that
θ v = θ v ε 0123 = ε μ 123 v ; 0 μ + ε 0 μ 23 v μ + ε 01 μ 3 v ; 2 μ + ε 012 μ v ; 3 μ = v ; μ μ . θ v = θ v ε 0123 = ε μ 123 v ; 0 μ + ε 0 μ 23 v μ + ε 01 μ 3 v ; 2 μ + ε 012 μ v ; 3 μ = v ; μ μ . {:[-theta_(v)=-theta_(v)epsi_(0123)=epsi_(mu123)v_(;0)^(mu)+epsi_(0mu23)v^(mu)],[+epsi_(01 mu3)v_(;2)^(mu)+epsi_(012 mu)v_(;3)^(mu)],[=v_(;mu)^(mu).]:}\begin{aligned} -\theta_{\boldsymbol{v}}=-\theta_{\boldsymbol{v}} \varepsilon_{0123}= & \varepsilon_{\mu 123} v_{; 0}^{\mu}+\varepsilon_{0 \mu 23} v^{\mu} \\ & +\varepsilon_{01 \mu 3} v_{; 2}^{\mu}+\varepsilon_{012 \mu} v_{; 3}^{\mu} \\ = & v_{; \mu}^{\mu} . \end{aligned}θv=θvε0123=εμ123v;0μ+ε0μ23vμ+ε01μ3v;2μ+ε012μv;3μ=v;μμ.
(37.6) (a) Trivially, £ J J = 0 £ J J = 0 £_(J)J=0£_{J} J=0£JJ=0. The other equalities follow from the fact that the box is rigid, so its corners must continue to be connected after they are transported along their geodesics.
(b) This follows from the result of the previous problem. (c) The quantity χ χ chi\chiχ is simply the number of particles in the box. Since this is a constant, by construction, we have £ J χ = 0 £ J χ = 0 £_(J)chi=0£_{J} \chi=0£Jχ=0.
(d) Since θ J = 0 θ J = 0 theta_(J)=0\theta_{\boldsymbol{J}}=0θJ=0, we also have from the previous problem that J = 0 J = 0 grad*J=0\boldsymbol{\nabla} \cdot \boldsymbol{J}=0J=0.
(38.2) (a) If ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~ is exact then ω ~ = d A ~ ω ~ = d A ~ tilde(omega)=d tilde(A)\tilde{\boldsymbol{\omega}}=\boldsymbol{d} \tilde{\boldsymbol{A}}ω~=dA~, where A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~ is a 1-form. By Stokes theorem
(E.360) S 2 ω ~ = S 2 d A ~ = S 2 A ~ = 0 (E.360) S 2 ω ~ = S 2 d A ~ = S 2 A ~ = 0 {:(E.360)int_(S^(2)) tilde(omega)=int_(S^(2))d tilde(A)=int_(delS^(2)) tilde(A)=0:}\begin{equation*} \int_{S^{2}} \tilde{\omega}=\int_{S^{2}} d \tilde{\boldsymbol{A}}=\int_{\partial S^{2}} \tilde{\boldsymbol{A}}=0 \tag{E.360} \end{equation*}(E.360)S2ω~=S2dA~=S2A~=0
since S 2 = 0 S 2 = 0 delS^(2)=0\partial S^{2}=0S2=0.
(b) We have d G ~ = d x 1 d x 2 d x 3 d G ~ = d x 1 d x 2 d x 3 d tilde(G)=dx^(1)^^dx^(2)^^dx^(3)\boldsymbol{d} \tilde{\boldsymbol{G}}=\boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}dG~=dx1dx2dx3, which is the volume 3 -form. Taken over the volume V V VVV of the ball from the last question, the integral
(E.361) V d G ~ 0 (E.361) V d G ~ 0 {:(E.361)int_(V)d tilde(G)!=0:}\begin{equation*} \int_{V} d \tilde{G} \neq 0 \tag{E.361} \end{equation*}(E.361)VdG~0
since it must give the volume inside the ball. By Stokes' theorem this tell us that
(E.362) S 2 G ~ 0 (E.362) S 2 G ~ 0 {:(E.362)int_(S^(2)) tilde(G)!=0:}\begin{equation*} \int_{S^{2}} \tilde{\boldsymbol{G}} \neq 0 \tag{E.362} \end{equation*}(E.362)S2G~0
Given the result from (a), this implies that G G GGG is not exact on S 2 S 2 S^(2)S^{2}S2. It is, however, closed on S 2 S 2 S^(2)S^{2}S2, as is the case for all 2 -forms, since all 3 -forms vanish in a two-dimensional space.
(38.4) (a) Using the results in Chapter 36 we find Γ ϕ ^ ϕ ^ θ ^ = cot θ Γ ϕ ^ ϕ ^ θ ^ = cot θ Gamma^( hat(phi))_( hat(phi) hat(theta))=cot theta\Gamma^{\hat{\phi}}{ }_{\hat{\phi} \hat{\theta}}=\cot \thetaΓϕ^ϕ^θ^=cotθ and Γ ϕ ˙ ϕ ^ ϕ ^ = cot θ Γ ϕ ˙ ϕ ^ ϕ ^ = cot θ Gamma_(phi^(˙) hat(phi))^( hat(phi))=-cot theta\Gamma_{\dot{\phi} \hat{\phi}}^{\hat{\phi}}=-\cot \thetaΓϕ˙ϕ^ϕ^=cotθ. We then use the geodesic equation, noting that u ϕ ^ = ( e ϕ ) ϕ ^ u ϕ = sin θ ϕ ˙ u ϕ ^ = e ϕ ϕ ^ u ϕ = sin θ ϕ ˙ u^( hat(phi))=(e_(phi))^( hat(phi))u^(phi)=sin thetaphi^(˙)u^{\hat{\phi}}=\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} u^{\phi}=\sin \theta \dot{\phi}uϕ^=(eϕ)ϕ^uϕ=sinθϕ˙.
(c) A trick is needed here: we write the integral I I I\mathcal{I}I as
I = 0 T d τ [ cos θ ( τ ) ϕ ˙ ( τ ) + 0 × θ ˙ ( τ ) ] (E.363) = 0 T [ cos θ d ϕ + 0 × d θ ] I = 0 T d τ [ cos θ ( τ ) ϕ ˙ ( τ ) + 0 × θ ˙ ( τ ) ] (E.363) = 0 T [ cos θ d ϕ + 0 × d θ ] {:[I=int_(0)^(T)dtau[cos theta(tau)phi^(˙)(tau)+0xxtheta^(˙)(tau)]],[(E.363)=int_(0)^(T)[cos thetadphi+0xxdtheta]]:}\begin{align*} \mathcal{I} & =\int_{0}^{T} \mathrm{~d} \tau[\cos \theta(\tau) \dot{\phi}(\tau)+0 \times \dot{\theta}(\tau)] \\ & =\int_{0}^{T}[\cos \theta \mathrm{~d} \phi+0 \times \mathrm{d} \theta] \tag{E.363} \end{align*}I=0T dτ[cosθ(τ)ϕ˙(τ)+0×θ˙(τ)](E.363)=0T[cosθ dϕ+0×dθ]
which we interpret as taken around a path, parametrized by τ τ tau\tauτ, that forms a closed loop that bounds a surface S S SSS. This allows us to use Green's theorem in the plane to write
I = S d ϕ d θ ( θ cos θ ϕ 0 ) (E.364) = S d ϕ d θ sin θ = A I = S d ϕ d θ θ cos θ ϕ 0 (E.364) = S d ϕ d θ sin θ = A {:[I=int_(S)dphidtheta(del_(theta)cos theta-del_(phi)0)],[(E.364)=-int_(S)dphidtheta sin theta=-A]:}\begin{align*} \mathcal{I} & =\int_{S} \mathrm{~d} \phi \mathrm{~d} \theta\left(\partial_{\theta} \cos \theta-\partial_{\phi} 0\right) \\ & =-\int_{S} \mathrm{~d} \phi \mathrm{~d} \theta \sin \theta=-\mathcal{A} \tag{E.364} \end{align*}I=S dϕ dθ(θcosθϕ0)(E.364)=S dϕ dθsinθ=A
This means we can write
(E.365) R = exp ( 0 A A 0 ) (E.365) R _ = exp 0 A A 0 {:(E.365)R_=exp([0,-A],[A,0]):}\underline{\boldsymbol{R}}=\exp \left(\begin{array}{cc} 0 & -\mathcal{A} \tag{E.365}\\ \mathcal{A} & 0 \end{array}\right)(E.365)R=exp(0AA0)
Expanding the exponential exp M = 1 + M + M 2 / 2 ! + exp M _ = 1 + M _ + M _ 2 / 2 ! + exp M_=1+M_+M_^(2)//2!+\exp \underline{\boldsymbol{M}}=1+\underline{\boldsymbol{M}}+\underline{\boldsymbol{M}}^{2} / 2!+expM=1+M+M2/2!+ ..., we find the rotation matrix
(E.366) R = ( cos A sin A sin A cos A ) (E.366) R _ = cos A sin A sin A cos A {:(E.366)R_=([cos A,-sin A],[sin A,cos A]):}\underline{\boldsymbol{R}}=\left(\begin{array}{cc} \cos \mathcal{A} & -\sin \mathcal{A} \tag{E.366}\\ \sin \mathcal{A} & \cos \mathcal{A} \end{array}\right)(E.366)R=(cosAsinAsinAcosA)
(39.1) Since ρ g y + 1 2 ρ v 2 + p ρ g y + 1 2 ρ v 2 + p rho gy+(1)/(2)rhov^(2)+p\rho g y+\frac{1}{2} \rho v^{2}+pρgy+12ρv2+p (where y y yyy is the height) is a constant, and the pressure is the same at the top of the reservoir as at the outlet. We have
(E.367) ρ g y 2 = ρ g y 1 + 1 2 ρ v 2 (E.367) ρ g y 2 = ρ g y 1 + 1 2 ρ v 2 {:(E.367)rho gy_(2)=rho gy_(1)+(1)/(2)*rhov^(2):}\begin{equation*} \rho g y_{2}=\rho g y_{1}+\frac{1}{2} \cdot \rho v^{2} \tag{E.367} \end{equation*}(E.367)ρgy2=ρgy1+12ρv2
since at the top of the reservoir (at height y 2 y 2 y_(2)y_{2}y2 ) we assume the fluid has no velocity. We obtain v = [ 2 g ( y 2 y 1 ) ] 1 2 v = 2 g y 2 y 1 1 2 v=[2g(y_(2)-y_(1))]^((1)/(2))v=\left[2 g\left(y_{2}-y_{1}\right)\right]^{\frac{1}{2}}v=[2g(y2y1)]12, which is the same result for a particle having fallen through a distance y 2 y 1 y 2 y 1 y_(2)-y_(1)y_{2}-y_{1}y2y1 in a constant gravitational field g g ggg.
(39.2) (a) We find
[ ( ρ + p ) u 0 u μ + p η 0 μ ] , μ = ( ρ + p ) , μ γ u μ + ( ρ + p ) γ , μ u μ + ( ρ + p ) γ u μ , μ + p , μ η 0 μ = ( ρ + p ) t γ u 0 + ( ρ + p ) x i γ u i + ( ρ + p ) γ t u 0 + ( ρ + p ) γ x i u i + ( ρ + p ) γ u 0 t + ( ρ + p ) γ u i x i p t = γ d d τ ( ρ + p ) + ( ρ + p ) d γ d τ + ( p + ρ ) γ u = d d τ [ γ ( ρ + p ) ] + ( p + ρ ) γ u = 0 , (E.368) ( ρ + p ) u 0 u μ + p η 0 μ , μ = ( ρ + p ) , μ γ u μ + ( ρ + p ) γ , μ u μ + ( ρ + p ) γ u μ , μ + p , μ η 0 μ = ( ρ + p ) t γ u 0 + ( ρ + p ) x i γ u i + ( ρ + p ) γ t u 0 + ( ρ + p ) γ x i u i + ( ρ + p ) γ u 0 t + ( ρ + p ) γ u i x i p t = γ d d τ ( ρ + p ) + ( ρ + p ) d γ d τ + ( p + ρ ) γ u = d d τ [ γ ( ρ + p ) ] + ( p + ρ ) γ u = 0 ,  (E.368)  {:[[(rho+p)u^(0)u^(mu)+peta^(0mu)]_(,mu)],[=(rho+p)_(,mu)gammau^(mu)+(rho+p)gamma","muu^(mu)],[+(rho+p)gammau^(mu)","mu+p","mueta^(0mu)],[=(del(rho+p))/(del t)gammau^(0)+(del(rho+p))/(delx^(i))gammau^(i)+(rho+p)(del gamma)/(del t)u^(0)],[+(rho+p)(del gamma)/(delx^(i))u^(i)+(rho+p)gamma(delu^(0))/(del t)],[+(rho+p)gamma(delu^(i))/(delx^(i))-(del p)/(del t)],[=gamma(d)/((d)tau)*(rho+p)+(rho+p)(dgamma)/((d)tau)+(p+rho)gamma vec(grad)* vec(u)],[=(d)/((d)tau)[gamma(rho+p)]+(p+rho)gamma vec(grad)* vec(u)=0","quad" (E.368) "]:}\begin{aligned} & {\left[(\rho+p) u^{0} u^{\mu}+p \eta^{0 \mu}\right]_{, \mu} } \\ = & (\rho+p)_{, \mu} \gamma u^{\mu}+(\rho+p) \gamma, \mu u^{\mu} \\ & +(\rho+p) \gamma u^{\mu}, \mu+p, \mu \eta^{0 \mu} \\ = & \frac{\partial(\rho+p)}{\partial t} \gamma u^{0}+\frac{\partial(\rho+p)}{\partial x^{i}} \gamma u^{i}+(\rho+p) \frac{\partial \gamma}{\partial t} u^{0} \\ & +(\rho+p) \frac{\partial \gamma}{\partial x^{i}} u^{i}+(\rho+p) \gamma \frac{\partial u^{0}}{\partial t} \\ & +(\rho+p) \gamma \frac{\partial u^{i}}{\partial x^{i}}-\frac{\partial p}{\partial t} \\ = & \gamma \frac{\mathrm{d}}{\mathrm{~d} \tau} \cdot(\rho+p)+(\rho+p) \frac{\mathrm{d} \gamma}{\mathrm{~d} \tau}+(p+\rho) \gamma \vec{\nabla} \cdot \vec{u} \\ = & \frac{\mathrm{d}}{\mathrm{~d} \tau}[\gamma(\rho+p)]+(p+\rho) \gamma \vec{\nabla} \cdot \vec{u}=0, \quad \text { (E.368) } \end{aligned}[(ρ+p)u0uμ+pη0μ],μ=(ρ+p),μγuμ+(ρ+p)γ,μuμ+(ρ+p)γuμ,μ+p,μη0μ=(ρ+p)tγu0+(ρ+p)xiγui+(ρ+p)γtu0+(ρ+p)γxiui+(ρ+p)γu0t+(ρ+p)γuixipt=γd dτ(ρ+p)+(ρ+p)dγ dτ+(p+ρ)γu=d dτ[γ(ρ+p)]+(p+ρ)γu=0, (E.368) 
where we've assumed steady flow.
(b) Start with the previous result and substitute using
conservation of rest mass
0 = d d τ [ γ ( ρ + p ) ] + ( p + ρ ) γ u = d d τ [ γ ( ρ + p ) ] ( p + ρ ) γ ρ 0 d ρ 0 d τ 0 = d d τ [ γ ( ρ + p ) ] + ( p + ρ ) γ u = d d τ [ γ ( ρ + p ) ] ( p + ρ ) γ ρ 0 d ρ 0 d τ {:[0=(d)/((d)tau)*[gamma(rho+p)]+(p+rho)gamma vec(grad)* vec(u)],[=(d)/((d)tau)*[gamma(rho+p)]-((p+rho)gamma)/(rho_(0))*((d)rho_(0))/((d)tau)]:}\begin{aligned} 0 & =\frac{\mathrm{d}}{\mathrm{~d} \tau} \cdot[\gamma(\rho+p)]+(p+\rho) \gamma \vec{\nabla} \cdot \vec{u} \\ & =\frac{\mathrm{d}}{\mathrm{~d} \tau} \cdot[\gamma(\rho+p)]-\frac{(p+\rho) \gamma}{\rho_{0}} \cdot \frac{\mathrm{~d} \rho_{0}}{\mathrm{~d} \tau} \end{aligned}0=d dτ[γ(ρ+p)]+(p+ρ)γu=d dτ[γ(ρ+p)](p+ρ)γρ0 dρ0 dτ
= d d τ [ γ ( ρ + p ) ρ 0 ] = γ v [ γ ( ρ + p ) ρ 0 ] , (E.369) = d d τ γ ( ρ + p ) ρ 0 = γ v γ ( ρ + p ) ρ 0 ,  (E.369)  =(d)/((d)tau)*[(gamma(rho+p))/(rho_(0))]=gamma vec(v)* vec(grad)[(gamma(rho+p))/(rho_(0))],quad" (E.369) "=\frac{\mathrm{d}}{\mathrm{~d} \tau} \cdot\left[\frac{\gamma(\rho+p)}{\rho_{0}}\right]=\gamma \vec{v} \cdot \vec{\nabla}\left[\frac{\gamma(\rho+p)}{\rho_{0}}\right], \quad \text { (E.369) }=d dτ[γ(ρ+p)ρ0]=γv[γ(ρ+p)ρ0], (E.369) 
where the final equality follows for the steady-flow condition d d τ = u = u 0 t + u = γ v d d τ = u = u 0 t + u = γ v (d)/(d tau)=u*grad=u^(0)(del)/(del t)+ vec(u)* vec(grad)=gamma vec(v)* vec(grad)\frac{d}{d \tau}=\boldsymbol{u} \cdot \nabla=u^{0} \frac{\partial}{\partial t}+\vec{u} \cdot \vec{\nabla}=\gamma \vec{v} \cdot \vec{\nabla}ddτ=u=u0t+u=γv.
(39.3) (a) Start with the conservation equation ( n u α ) , α = 0 n u α , α = 0 (nu^(alpha)),alpha=0\left(n u^{\alpha}\right), \alpha=0(nuα),α=0, where we use commas rather than semicolons since we're working in flat spacetime. We obtain
x α [ ( n + δ n ) ( u α + δ u α ] = 0 x α n + δ n u α + δ u α = 0 (del)/(delx^(alpha))[(n_(**)+delta n)(u_(**)^(alpha)+deltau^(alpha)]=0:}\frac{\partial}{\partial x^{\alpha}}\left[\left(n_{*}+\delta n\right)\left(u_{*}^{\alpha}+\delta u^{\alpha}\right]=0\right.xα[(n+δn)(uα+δuα]=0
For the first-order contribution, we only need consider
x α [ ( u α δ n + n δ u α ] = 0 x α u α δ n + n δ u α = 0 (del)/(delx^(alpha))[(u_(**)^(alpha)delta n+n_(**)deltau^(alpha)]=0:}\frac{\partial}{\partial x^{\alpha}}\left[\left(u_{*}^{\alpha} \delta n+n_{*} \delta u^{\alpha}\right]=0\right.xα[(uαδn+nδuα]=0
Plugging in components yields the desired expression. (b) Start with the Euler equation in the form ( ρ + ρ + rho+\rho+ρ+ p) u α , ν u ν + p , α + p , ν u ν u α = 0 u α , ν u ν + p , α + p , ν u ν u α = 0 u_(alpha,nu)u^(nu)+p_(,alpha)+p_(,nu)u^(nu)u_(alpha)=0u_{\alpha, \nu} u^{\nu}+p_{, \alpha}+p_{, \nu} u^{\nu} u_{\alpha}=0uα,νuν+p,α+p,νuνuα=0, and repeat the same procedure as in part (a) to obtain the equation.
(c) If we have for a perfect fluid that
0 = d s d τ = u α s , α (E.372) = ( t + u i i ) s = t δ s , 0 = d s d τ = u α s , α (E.372) = t + u i i s = t δ s , {:[0=(ds)/((d)tau)=u^(alpha)s","alpha],[(E.372)=((del)/(del t)+u^(i)grad_(i))s=(del)/(del t)delta s","]:}\begin{align*} 0=\frac{\mathrm{d} s}{\mathrm{~d} \tau} & =u^{\alpha} s, \alpha \\ & =\left(\frac{\partial}{\partial t}+u^{i} \nabla_{i}\right) s=\frac{\partial}{\partial t} \delta s, \tag{E.372} \end{align*}0=ds dτ=uαs,α(E.372)=(t+uii)s=tδs,
then we have an unchanging entropy to first order. Plugging into the first law of thermodynamics then yields the stated equation.
(d) Using the definition, we can write δ p = c s 2 δ ρ δ p = c s 2 δ ρ delta p=c_(s)^(2)delta rho\delta p=c_{\mathrm{s}}^{2} \delta \rhoδp=cs2δρ, so that the first law gives c s 2 ( p + ρ ) δ n / n = δ p c s 2 p + ρ δ n / n = δ p c_(s)^(2)(p_(**)+rho_(**))delta n//n_(**)=delta pc_{s}^{2}\left(p_{*}+\rho_{*}\right) \delta n / n_{*}=\delta pcs2(p+ρ)δn/n=δp. Substituting into the result from (b) we find
(E.373) t δ v + c s 2 n δ n = 0 (E.373) t δ v + c s 2 n δ n = 0 {:(E.373)(del)/(del t)delta vec(v)+(c_(s)^(2))/(n_(**)) vec(grad)delta n=0:}\begin{equation*} \frac{\partial}{\partial t} \delta \vec{v}+\frac{c_{s}^{2}}{n_{*}} \vec{\nabla} \delta n=0 \tag{E.373} \end{equation*}(E.373)tδv+cs2nδn=0
Take the gradient of this latter expression and substitute into the result from (a) to obtain the wave equation.
(39.4) Using the identities given in the question we have
J m = m g d τ d z α d τ x α δ ( 4 ) [ x z ( τ ) ] J m = m g d τ d z α d τ x α δ ( 4 ) [ x z ( τ ) ] grad*J_(m)=(m)/(sqrt(-g))intdtau((d)z^(alpha))/(dtau)(del)/(delx^(alpha))delta^((4))[x-z(tau)]\boldsymbol{\nabla} \cdot \boldsymbol{J}_{\mathrm{m}}=\frac{m}{\sqrt{-g}} \int \mathrm{~d} \tau \frac{\mathrm{~d} z^{\alpha}}{\mathrm{d} \tau} \frac{\partial}{\partial x^{\alpha}} \delta^{(4)}[x-z(\tau)]Jm=mg dτ dzαdτxαδ(4)[xz(τ)]
= m g d τ d d τ δ ( 4 ) [ x z ( τ ) ] = 0 = m g d τ d d τ δ ( 4 ) [ x z ( τ ) ] = 0 =-(m)/(sqrt(-g))intdtau((d))/((d)tau)delta^((4))[x-z(tau)]=0=-\frac{m}{\sqrt{-g}} \int \mathrm{~d} \tau \frac{\mathrm{~d}}{\mathrm{~d} \tau} \delta^{(4)}[x-z(\tau)]=0=mg dτ d dτδ(4)[xz(τ)]=0
(39.5) The only components of T T T\boldsymbol{T}T we need are T t t = ρ e 2 Φ T t t = ρ e 2 Φ T^(tt)=rhoe^(-2Phi)T^{t t}=\rho \mathrm{e}^{-2 \Phi}Ttt=ρe2Φ and T r r = p e 2 Λ T r r = p e 2 Λ T^(rr)=pe^(-2Lambda)T^{r r}=p \mathrm{e}^{-2 \Lambda}Trr=pe2Λ. We consider the component equation
T μ ν ; μ = T μ ν , μ + Γ μ μ σ T σ ν + Γ ν μ σ T μ σ . T μ ν ; μ = T μ ν , μ + Γ μ μ σ T σ ν + Γ ν μ σ T μ σ . T^(mu nu)_(;mu)=T^(mu nu)_(,mu)+Gamma^(mu)_(mu sigma)T^(sigma nu)+Gamma^(nu)_(mu sigma)T^(mu sigma).quadT^{\mu \nu}{ }_{; \mu}=T^{\mu \nu}{ }_{, \mu}+\Gamma^{\mu}{ }_{\mu \sigma} T^{\sigma \nu}+\Gamma^{\nu}{ }_{\mu \sigma} T^{\mu \sigma} . \quadTμν;μ=Tμν,μ+ΓμμσTσν+ΓνμσTμσ. (E.375)
First set μ = t μ = t mu=t\mu=tμ=t and ν = r ν = r nu=r\nu=rν=r to find
T ; t t r = 0 + Γ t r t T r r + Γ t t r T t t T ; t t r = 0 + Γ t r t T r r + Γ t t r T t t T_(;t)^(tr)=0+Gamma_(tr)^(t)T^(rr)+Gamma_(tt)^(r)T^(tt)T_{; t}^{t r}=0+\Gamma_{t r}^{t} T^{r r}+\Gamma_{t t}^{r} T^{t t}T;ttr=0+ΓtrtTrr+ΓttrTtt
(E.376) = Φ ( p + ρ ) e 2 Λ (E.376) = Φ ( p + ρ ) e 2 Λ {:(E.376)=Phi^(')(p+rho)e^(-2Lambda):}\begin{equation*} =\Phi^{\prime}(p+\rho) \mathrm{e}^{-2 \Lambda} \tag{E.376} \end{equation*}(E.376)=Φ(p+ρ)e2Λ
Then set μ = r μ = r mu=r\mu=rμ=r and ν = r ν = r nu=r\nu=rν=r to find
T r r ; r = T r r r + Γ r r r T r r + Γ r r r T r r T r r ; r = T r r r + Γ r r r T r r + Γ r r r T r r T^(rr)_(;r)=(delT^(rr))/(del r)+Gamma^(r)_(rr)T^(rr)+Gamma^(r)_(rr)T^(rr)T^{r r}{ }_{; r}=\frac{\partial T^{r r}}{\partial r}+\Gamma^{r}{ }_{r r} T^{r r}+\Gamma^{r}{ }_{r r} T^{r r}Trr;r=Trrr+ΓrrrTrr+ΓrrrTrr
= ( p e 2 Λ ) r + 2 Λ p e 2 Λ = p e 2 Λ r + 2 Λ p e 2 Λ =(del(pe^(-2Lambda)))/(del r)+2Lambda^(')pe^(-2Lambda)=\frac{\partial\left(p \mathrm{e}^{-2 \Lambda}\right)}{\partial r}+2 \Lambda^{\prime} p \mathrm{e}^{-2 \Lambda}=(pe2Λ)r+2Λpe2Λ
(E.377) = p r e 2 Λ (E.377) = p r e 2 Λ {:(E.377)=(del p)/(del r)e^(-2Lambda):}\begin{equation*} =\frac{\partial p}{\partial r} \mathrm{e}^{-2 \Lambda} \tag{E.377} \end{equation*}(E.377)=pre2Λ
Adding the two contributions and equating them to zero then gives the answer.
(39.6) (b) We first note that d ω ~ , v = 0 d ω ~ , v = 0 (:d tilde(omega), vec(v):)=0\langle\boldsymbol{d} \tilde{\omega}, \vec{v}\rangle=0dω~,v=0. Then, plugging in, (40.4) Computing we find
we have
we have £ v ( ρ 0 ω ~ ) = d ω ~ , ρ 0 v = d ( ρ 0 v x d y d z ρ 0 v y d x d z + ρ 0 v z d x d y ) = ( ( ρ 0 v x ) x + ( ρ 0 v y ) y + ( ρ 0 v z ) z ) d x d y d z (E.378) = [ ( ρ 0 v ) ] ω ~ .  we have  £ v ρ 0 ω ~ = d ω ~ , ρ 0 v = d ρ 0 v x d y d z ρ 0 v y d x d z + ρ 0 v z d x d y = ρ 0 v x x + ρ 0 v y y + ρ 0 v z z d x d y d z (E.378) = ρ 0 v ω ~ . {:[" we have "],[£_( vec(v))(rho_(0)( tilde(omega)))],[=d(:( tilde(omega)),rho_(0)( vec(v)):)],[=d(rho_(0)v^(x)dy^^dz-rho_(0)v^(y)dx^^dz+rho_(0)v^(z)dx^^dy)],[=((del(rho_(0)v^(x)))/(del x)+(del(rho_(0)v^(y)))/(del y)+(del(rho_(0)v^(z)))/(del z))dx^^dy^^dz],[(E.378)=[( vec(grad))*(rho_(0)( vec(v)))] tilde(omega).]:}\begin{align*} & \text { we have } \\ & £_{\vec{v}}\left(\rho_{0} \tilde{\boldsymbol{\omega}}\right) \\ &=\boldsymbol{d}\left\langle\tilde{\boldsymbol{\omega}}, \rho_{0} \vec{v}\right\rangle \\ &= \boldsymbol{d}\left(\rho_{0} v^{x} \boldsymbol{d} y \wedge \boldsymbol{d} z-\rho_{0} v^{y} \boldsymbol{d} x \wedge \boldsymbol{d} z+\rho_{0} v^{z} \boldsymbol{d} x \wedge \boldsymbol{d} y\right) \\ &=\left(\frac{\partial\left(\rho_{0} v^{x}\right)}{\partial x}+\frac{\partial\left(\rho_{0} v^{y}\right)}{\partial y}+\frac{\partial\left(\rho_{0} v^{z}\right)}{\partial z}\right) \boldsymbol{d} x \wedge \boldsymbol{d} y \wedge \boldsymbol{d} z \\ &= {\left[\vec{\nabla} \cdot\left(\rho_{0} \vec{v}\right)\right] \tilde{\boldsymbol{\omega}} . } \tag{E.378} \end{align*} we have £v(ρ0ω~)=dω~,ρ0v=d(ρ0vxdydzρ0vydxdz+ρ0vzdxdy)=((ρ0vx)x+(ρ0vy)y+(ρ0vz)z)dxdydz(E.378)=[(ρ0v)]ω~.
(39.7) Since v j = v j v j = v j v_(j)=v^(j)v_{j}=v^{j}vj=vj in flat space, we can write a 1 -form version of the given equation as
( £ v ~ v ~ ) i = v j x j v i + v j x i v j (E.379) = v j x j v i + 1 2 x i ( v j v j ) £ v ~ v ~ i = v j x j v i + v j x i v j (E.379) = v j x j v i + 1 2 x i v j v j {:[(£_( tilde(v))( tilde(v)))_(i)=v^(j)(del)/(delx^(j))v_(i)+v_(j)(del)/(delx^(i))v^(j)],[(E.379)=v^(j)(del)/(delx^(j))v_(i)+(1)/(2)(del)/(delx^(i))(v_(j)v^(j))]:}\begin{align*} \left(£_{\tilde{v}} \tilde{v}\right)_{i} & =v^{j} \frac{\partial}{\partial x^{j}} v_{i}+v_{j} \frac{\partial}{\partial x^{i}} v^{j} \\ & =v^{j} \frac{\partial}{\partial x^{j}} v_{i}+\frac{1}{2} \frac{\partial}{\partial x^{i}}\left(v_{j} v^{j}\right) \tag{E.379} \end{align*}(£v~v~)i=vjxjvi+vjxivj(E.379)=vjxjvi+12xi(vjvj)
and we have
(E.380) v j x j v i = ( £ v v ~ ) i x i ( v 2 / 2 ) . (E.380) v j x j v i = £ v v ~ i x i v 2 / 2 . {:(E.380)v^(j)(del)/(delx^(j))v_(i)=(£_( vec(v))( tilde(v)))_(i)-(del)/(delx^(i))(v^(2)//2).:}\begin{equation*} v^{j} \frac{\partial}{\partial x^{j}} v_{i}=\left(£_{\vec{v}} \tilde{\boldsymbol{v}}\right)_{i}-\frac{\partial}{\partial x^{i}}\left(v^{2} / 2\right) . \tag{E.380} \end{equation*}(E.380)vjxjvi=(£vv~)ixi(v2/2).
Substituting the other derivatives for ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) exterior derivatives for consistency, we get the answer.
(39.8) A suitable 2-form is f ~ = ε i j k v i [ d x j ( ) d x k ( ) ] f ~ = ε i j k v i d x j ( ) d x k ( ) tilde(f)=epsi_(ijk)v^(i)[dx^(j)()^^dx^(k)()]\tilde{\boldsymbol{f}}=\varepsilon_{i j k} v^{i}\left[\boldsymbol{d} x^{j}() \wedge \boldsymbol{d} x^{k}()\right]f~=εijkvi[dxj()dxk()], where ε i j k ε i j k epsi_(ijk)\varepsilon_{i j k}εijk is the three-dimensional antisymmetric symbol. When the two displacement vectors are inserted into the slots, we obtain the triple scalar product between v , a v , a vec(v), vec(a)\vec{v}, \vec{a}v,a and b b vec(b)\vec{b}b.
(40.1) (a) Using the Euler-Lagrange equation, the equation of motion is found to be
(E.381) d d τ ( g μ ν x ˙ ν ) = 1 2 g α β x μ x ˙ α x ˙ β (E.381) d d τ g μ ν x ˙ ν = 1 2 g α β x μ x ˙ α x ˙ β {:(E.381)(d)/((d)tau)(g_(mu nu)x^(˙)^(nu))=(1)/(2)(delg_(alpha beta))/(delx^(mu))x^(˙)^(alpha)x^(˙)^(beta):}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} \tau}\left(g_{\mu \nu} \dot{x}^{\nu}\right)=\frac{1}{2} \frac{\partial g_{\alpha \beta}}{\partial x^{\mu}} \dot{x}^{\alpha} \dot{x}^{\beta} \tag{E.381} \end{equation*}(E.381)d dτ(gμνx˙ν)=12gαβxμx˙αx˙β
The geodesic equation, in its usual form, follows from the steps used in the derivation in Chapter 9.
(c) The equations of motion are
x ˙ μ = g μ ν p ν , p ˙ μ = 1 2 g α β x μ p α p β x ˙ μ = g μ ν p ν , p ˙ μ = 1 2 g α β x μ p α p β x^(˙)^(mu)=g^(mu nu)p_(nu),quadp^(˙)_(mu)=-(1)/(2)(delg^(alpha beta))/(delx^(mu))*p_(alpha)p_(beta)\dot{x}^{\mu}=g^{\mu \nu} p_{\nu}, \quad \dot{p}_{\mu}=-\frac{1}{2} \frac{\partial g^{\alpha \beta}}{\partial x^{\mu}} \cdot p_{\alpha} p_{\beta}x˙μ=gμνpν,p˙μ=12gαβxμpαpβ
These can be used to derive the geodesic equation in the form given in (a), noting the minus sign going between δ g μ ν δ g μ ν deltag^(mu nu)\delta g^{\mu \nu}δgμν and δ g μ ν δ g μ ν deltag_(mu nu)\delta g_{\mu \nu}δgμν.
(d) A metric component with no dependence on x σ x σ x^(sigma)x^{\sigma}xσ causes the right-hand side of the equation of motion to vanish, showing that momentum component p σ p σ p_(sigma)p_{\sigma}pσ is conserved.
(40.2) (a) We find
p t = ( 1 2 M r ) t ˙ , p r = ( 1 2 M r ) 1 r ˙ , p θ = r 2 θ ˙ , p ϕ = r 2 sin 2 θ ϕ ˙ . p t = 1 2 M r t ˙ , p r = 1 2 M r 1 r ˙ , p θ = r 2 θ ˙ , p ϕ = r 2 sin 2 θ ϕ ˙ . {:[p_(t)=-(1-(2M)/(r))t^(˙)",",p_(r)=(1-(2M)/(r))^(-1)r^(˙)","],[,p_(theta)=r^(2)theta^(˙)",",p_(phi)=r^(2)sin^(2)thetaphi^(˙).]:}\begin{array}{ccc} p_{t}=-\left(1-\frac{2 M}{r}\right) \dot{t}, & p_{r}=\left(1-\frac{2 M}{r}\right)^{-1} \dot{r}, \\ & p_{\theta}=r^{2} \dot{\theta}, & p_{\phi}=r^{2} \sin ^{2} \theta \dot{\phi} . \end{array}pt=(12Mr)t˙,pr=(12Mr)1r˙,pθ=r2θ˙,pϕ=r2sin2θϕ˙.
(b) We have that
(E.383) H t = H ϕ = 0 (E.383) H t = H ϕ = 0 {:(E.383)(del H)/(del t)=(del H)/(del phi)=0:}\begin{equation*} \frac{\partial H}{\partial t}=\frac{\partial H}{\partial \phi}=0 \tag{E.383} \end{equation*}(E.383)Ht=Hϕ=0
implying that both p t p t p_(t)p_{t}pt and p ϕ p ϕ p_(phi)p_{\phi}pϕ are constants of the motion.
M μ ν σ x σ = δ μ σ T ν σ + x μ T ν σ , σ δ ν σ T μ σ x ν T μ σ , σ . Since T μ ν , μ = 0 we obtain M μ ν σ x σ = T ν μ T μ ν = 0 , M μ ν σ x σ = δ μ σ T ν σ + x μ T ν σ , σ δ ν σ T μ σ x ν T μ σ , σ .  Since  T μ ν , μ = 0  we obtain  M μ ν σ x σ = T ν μ T μ ν = 0 , {:[(delM^(mu nu sigma))/(delx^(sigma))=delta^(mu)_(sigma)T^(nu sigma)+x^(mu)T^(nu sigma)_(,sigma)-delta^(nu)_(sigma)T^(mu sigma)-x^(nu)T^(mu sigma)_(,sigma).],[" Since "T^(mu nu)_(,mu)=0" we obtain "],[qquad(delM^(mu nu sigma))/(delx^(sigma))=T^(nu mu)-T^(mu nu)=0","]:}\begin{aligned} & \frac{\partial M^{\mu \nu \sigma}}{\partial x^{\sigma}}=\delta^{\mu}{ }_{\sigma} T^{\nu \sigma}+x^{\mu} T^{\nu \sigma}{ }_{, \sigma}-\delta^{\nu}{ }_{\sigma} T^{\mu \sigma}-x^{\nu} T^{\mu \sigma}{ }_{, \sigma} . \\ & \text { Since } T^{\mu \nu}{ }_{, \mu}=0 \text { we obtain } \\ & \qquad \frac{\partial M^{\mu \nu \sigma}}{\partial x^{\sigma}}=T^{\nu \mu}-T^{\mu \nu}=0, \end{aligned}Mμνσxσ=δμσTνσ+xμTνσ,σδνσTμσxνTμσ,σ. Since Tμν,μ=0 we obtain Mμνσxσ=TνμTμν=0,
which requires T T T\boldsymbol{T}T to be symmetrical.
(41.1) (a) We can compute
Φ ; μ ν = ( μ Φ ) ; ν = ν μ Φ Γ ν μ σ σ Φ Φ ; μ ν = μ Φ ; ν = ν μ Φ Γ ν μ σ σ Φ Phi_(;mu nu)=(del_(mu)Phi)_(;nu)=del_(nu)del_(mu)Phi-Gamma_(nu mu)^(sigma)del_(sigma)Phi\Phi_{; \mu \nu}=\left(\partial_{\mu} \Phi\right)_{; \nu}=\partial_{\nu} \partial_{\mu} \Phi-\Gamma_{\nu \mu}^{\sigma} \partial_{\sigma} \PhiΦ;μν=(μΦ);ν=νμΦΓνμσσΦ
using the connection coefficients. We obtain diagonal components
Φ ; t t = t 2 Φ Φ ; t t = t 2 Φ Phi_(;tt)=del_(t)^(2)Phi\Phi_{; t t}=\partial_{t}^{2} \PhiΦ;tt=t2Φ,
Φ ; r r = r 2 Φ a a ˙ 1 k r 2 t Φ k r 1 k r 2 r Φ Φ ; r r = r 2 Φ a a ˙ 1 k r 2 t Φ k r 1 k r 2 r Φ Phi_(;rr)=del_(r)^(2)Phi-(a(a^(˙)))/(1-kr^(2))del_(t)Phi-(kr)/(1-kr^(2))del_(r)Phi\Phi_{; r r}=\partial_{r}^{2} \Phi-\frac{a \dot{a}}{1-k r^{2}} \partial_{t} \Phi-\frac{k r}{1-k r^{2}} \partial_{r} \PhiΦ;rr=r2Φaa˙1kr2tΦkr1kr2rΦ,
Φ ; θ θ = θ 2 Φ a a ˙ r 2 t Φ + r ( 1 k r 2 ) r Φ Φ ; θ θ = θ 2 Φ a a ˙ r 2 t Φ + r 1 k r 2 r Φ Phi_(;theta theta)=del_(theta)^(2)Phi-aa^(˙)r^(2)del_(t)Phi+r(1-kr^(2))del_(r)Phi\Phi_{; \theta \theta}=\partial_{\theta}^{2} \Phi-a \dot{a} r^{2} \partial_{t} \Phi+r\left(1-k r^{2}\right) \partial_{r} \PhiΦ;θθ=θ2Φaa˙r2tΦ+r(1kr2)rΦ,
Φ ; ϕ ϕ = ϕ 2 Φ a a ˙ r 2 sin 2 θ t Φ + r ( 1 k r 2 ) sin 2 θ r Φ Φ ; ϕ ϕ = ϕ 2 Φ a a ˙ r 2 sin 2 θ t Φ + r 1 k r 2 sin 2 θ r Φ Phi_(;phi phi)=del_(phi)^(2)Phi-aa^(˙)r^(2)sin^(2)thetadel_(t)Phi+r(1-kr^(2))sin^(2)thetadel_(r)Phi\Phi_{; \phi \phi}=\partial_{\phi}^{2} \Phi-a \dot{a} r^{2} \sin ^{2} \theta \partial_{t} \Phi+r\left(1-k r^{2}\right) \sin ^{2} \theta \partial_{r} \PhiΦ;ϕϕ=ϕ2Φaa˙r2sin2θtΦ+r(1kr2)sin2θrΦ
+ sin θ cos θ θ Φ + sin θ cos θ θ Φ +sin theta cos thetadel_(theta)Phi+\sin \theta \cos \theta \partial_{\theta} \Phi+sinθcosθθΦ
(E.387)
These give us
g μ ν Φ ; μ ν = t 2 Φ + 1 k 2 r 2 a 2 ( r 2 Φ ( a a ˙ t Φ + k r r Φ ) 1 k r 2 ) + 1 a 2 r 2 [ θ 2 Φ a a ˙ r 2 t Φ + r ( 1 k r 2 ) r Φ ] + 1 a 2 r 2 sin 2 θ [ ϕ 2 Φ a a ˙ r 2 sin 2 θ t Φ + r ( 1 k r 2 ) sin 2 θ r Φ + sin θ cos θ θ Φ ] . (Е.388) g μ ν Φ ; μ ν = t 2 Φ + 1 k 2 r 2 a 2 r 2 Φ a a ˙ t Φ + k r r Φ 1 k r 2 + 1 a 2 r 2 θ 2 Φ a a ˙ r 2 t Φ + r 1 k r 2 r Φ + 1 a 2 r 2 sin 2 θ ϕ 2 Φ a a ˙ r 2 sin 2 θ t Φ + r 1 k r 2 sin 2 θ r Φ + sin θ cos θ θ Φ .  (Е.388)  {:[g^(mu nu)Phi_(;mu nu)=-del_(t)^(2)Phi+(1-k^(2)r^(2))/(a^(2))*(del_(r)^(2)Phi-((a(a^(˙))del_(t)Phi+krdel_(r)Phi))/(1-kr^(2)))],[+(1)/(a^(2)r^(2))*[del_(theta)^(2)Phi-a(a^(˙))r^(2)del_(t)Phi+r(1-kr^(2))del_(r)Phi]],[+(1)/(a^(2)r^(2)sin^(2)theta)[del_(phi)^(2)Phi-a(a^(˙))r^(2)sin^(2)thetadel_(t)Phi+:}],[{:r(1-kr^(2))sin^(2)thetadel_(r)Phi+sin theta cos thetadel_(theta)Phi].],[" (Е.388) "]:}\begin{aligned} & g^{\mu \nu} \Phi_{; \mu \nu}=-\partial_{t}^{2} \Phi+\frac{1-k^{2} r^{2}}{a^{2}} \cdot\left(\partial_{r}^{2} \Phi-\frac{\left(a \dot{a} \partial_{t} \Phi+k r \partial_{r} \Phi\right)}{1-k r^{2}}\right) \\ &+\frac{1}{a^{2} r^{2}} \cdot\left[\partial_{\theta}^{2} \Phi-a \dot{a} r^{2} \partial_{t} \Phi+r\left(1-k r^{2}\right) \partial_{r} \Phi\right] \\ &+\frac{1}{a^{2} r^{2} \sin ^{2} \theta}\left[\partial_{\phi}^{2} \Phi-a \dot{a} r^{2} \sin ^{2} \theta \partial_{t} \Phi+\right. \\ &\left.r\left(1-k r^{2}\right) \sin ^{2} \theta \partial_{r} \Phi+\sin \theta \cos \theta \partial_{\theta} \Phi\right] . \\ & \text { (Е.388) } \end{aligned}gμνΦ;μν=t2Φ+1k2r2a2(r2Φ(aa˙tΦ+krrΦ)1kr2)+1a2r2[θ2Φaa˙r2tΦ+r(1kr2)rΦ]+1a2r2sin2θ[ϕ2Φaa˙r2sin2θtΦ+r(1kr2)sin2θrΦ+sinθcosθθΦ]. (Е.388) 
Collecting terms we obtain the equation of motion
t 2 Φ + 1 k 2 r 2 a 2 r 2 Φ + 1 a 2 r 2 θ 2 Φ + 1 a 2 r 2 sin 2 θ ϕ 2 Φ 3 a ˙ a t Φ + [ ( 2 3 k r 2 ) a 2 r ] r Φ + cos θ a 2 r 2 sin θ θ Φ (E.389) + U ϕ = 0 t 2 Φ + 1 k 2 r 2 a 2 r 2 Φ + 1 a 2 r 2 θ 2 Φ + 1 a 2 r 2 sin 2 θ ϕ 2 Φ 3 a ˙ a t Φ + 2 3 k r 2 a 2 r r Φ + cos θ a 2 r 2 sin θ θ Φ (E.389) + U ϕ = 0 {:[-del_(t)^(2)Phi+(1-k^(2)r^(2))/(a^(2))del_(r)^(2)Phi+(1)/(a^(2)r^(2))del_(theta)^(2)Phi+(1)/(a^(2)r^(2)sin^(2)theta)del_(phi)^(2)Phi],[-3((a^(˙)))/(a)del_(t)Phi+[((2-3kr^(2)))/(a^(2)r)]del_(r)Phi+(cos theta)/(a^(2)r^(2)sin theta)del_(theta)Phi],[(E.389)+(del U)/(del phi)=0]:}\begin{align*} & -\partial_{t}^{2} \Phi+\frac{1-k^{2} r^{2}}{a^{2}} \partial_{r}^{2} \Phi+\frac{1}{a^{2} r^{2}} \partial_{\theta}^{2} \Phi+\frac{1}{a^{2} r^{2} \sin ^{2} \theta} \partial_{\phi}^{2} \Phi \\ & -3 \frac{\dot{a}}{a} \partial_{t} \Phi+\left[\frac{\left(2-3 k r^{2}\right)}{a^{2} r}\right] \partial_{r} \Phi+\frac{\cos \theta}{a^{2} r^{2} \sin \theta} \partial_{\theta} \Phi \\ & +\frac{\partial U}{\partial \phi}=0 \tag{E.389} \end{align*}t2Φ+1k2r2a2r2Φ+1a2r2θ2Φ+1a2r2sin2θϕ2Φ3a˙atΦ+[(23kr2)a2r]rΦ+cosθa2r2sinθθΦ(E.389)+Uϕ=0
(b) We can compare the previous expression against the rule for computing 2 2 vec(grad)^(2)\vec{\nabla}^{2}2 in spherical polars, which is 2 r 2 + 1 r 2 2 θ 2 + 1 r 2 sin 2 θ 2 ϕ 2 + 2 r r + cos θ r 2 sin θ ( E .390 ) 2 r 2 + 1 r 2 2 θ 2 + 1 r 2 sin 2 θ 2 ϕ 2 + 2 r r + cos θ r 2 sin θ ( E .390 ) (del^(2))/(delr^(2))+(1)/(r^(2))*(del^(2))/(deltheta^(2))+(1)/(r^(2)sin^(2)theta)*(del^(2))/(delphi^(2))+(2)/(r)(del)/(del r)+(cos theta)/(r^(2)sin theta)(del)/((E.390))\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r^{2}} \cdot \frac{\partial^{2}}{\partial \theta^{2}}+\frac{1}{r^{2} \sin ^{2} \theta} \cdot \frac{\partial^{2}}{\partial \phi^{2}}+\frac{2}{r} \frac{\partial}{\partial r}+\frac{\cos \theta}{r^{2} \sin \theta} \frac{\partial}{(\mathrm{E} .390)}2r2+1r22θ2+1r2sin2θ2ϕ2+2rr+cosθr2sinθ(E.390).
This allows us to check that the spatial parts behave as expected when k = a ˙ = 0 k = a ˙ = 0 k=a^(˙)=0k=\dot{a}=0k=a˙=0. If we are interested in slowly varying solutions in space, then the value of k k kkk does not varying solut
(42.2) For the usual passive transformation using eqn 2.10 the components are
( E ) x = E x ( E ) y = γ ( E y β B z ) ( E ) z = γ ( E z + β B y ) ( B ) x = B x ( B ) y = γ ( B y + β E z ) , (E.391) ( B ) z = γ ( B z β E y ) . E x = E x E y = γ E y β B z E z = γ E z + β B y B x = B x B y = γ B y + β E z , (E.391) B z = γ B z β E y . {:[(E^('))^(x)=E^(x)],[(E^('))^(y)=gamma(E^(y)-betaB^(z))],[(E^('))^(z)=gamma(E^(z)+betaB^(y))],[(B^('))^(x)=B^(x)],[(B^('))^(y)=gamma(B^(y)+betaE^(z))","],[(E.391)(B^('))^(z)=gamma(B^(z)-betaE^(y)).]:}\begin{align*} & \left(E^{\prime}\right)^{x}=E^{x} \\ & \left(E^{\prime}\right)^{y}=\gamma\left(E^{y}-\beta B^{z}\right) \\ & \left(E^{\prime}\right)^{z}=\gamma\left(E^{z}+\beta B^{y}\right) \\ & \left(B^{\prime}\right)^{x}=B^{x} \\ & \left(B^{\prime}\right)^{y}=\gamma\left(B^{y}+\beta E^{z}\right), \\ & \left(B^{\prime}\right)^{z}=\gamma\left(B^{z}-\beta E^{y}\right) . \tag{E.391} \end{align*}(E)x=Ex(E)y=γ(EyβBz)(E)z=γ(Ez+βBy)(B)x=Bx(B)y=γ(By+βEz),(E.391)(B)z=γ(BzβEy).
For an active transformation, reverse the sign of β β beta\betaβ.
(42.4) For both (a) and (b) we have B = C e z B = C e z vec(B)=C vec(e)_(z)\vec{B}=C \vec{e}_{z}B=Cez.
(42.13) (a) We find
(E.392) L m c 2 + 1 2 m v 2 λ ϕ + (E.392) L m c 2 + 1 2 m v 2 λ ϕ + {:(E.392)L~~-mc^(2)+(1)/(2)mv^(2)-lambda phi+dots:}\begin{equation*} L \approx-m c^{2}+\frac{1}{2} m v^{2}-\lambda \phi+\ldots \tag{E.392} \end{equation*}(E.392)Lmc2+12mv2λϕ+
(c) Invoking the equivalence principle for the equation of motion must remove the dependence on m m mmm. Scaling λ λ lambda\lambdaλ and rescaling the field, we find that the equation of motion must take the form
(E.393) d u μ d τ = 1 ( 1 + Φ ) ( Φ , μ + u μ u σ Φ σ ) (E.393) d u μ d τ = 1 ( 1 + Φ ) Φ , μ + u μ u σ Φ σ {:(E.393)(du_(mu))/(dtau)=-(1)/((1+Phi))*(Phi_(,mu)+u_(mu)u^(sigma)Phi_(sigma)):}\begin{equation*} \frac{\mathrm{d} u_{\mu}}{\mathrm{d} \tau}=-\frac{1}{(1+\Phi)} \cdot\left(\Phi_{, \mu}+u_{\mu} u^{\sigma} \Phi_{\sigma}\right) \tag{E.393} \end{equation*}(E.393)duμdτ=1(1+Φ)(Φ,μ+uμuσΦσ)
The action then takes the form suggested in the question for consistency.
(42.14) (b) A scalar that describes both matter and energy is the trace T T TTT of the energy-momentum tensor T T T\boldsymbol{T}T. Using this we can upgrade the Lagrangian to read
(E.394) L = 1 8 π G ( μ Φ ) 2 + T Φ (E.394) L = 1 8 π G μ Φ 2 + T Φ {:(E.394)L=-(1)/(8pi G)(del_(mu)Phi)^(2)+T Phi:}\begin{equation*} \mathcal{L}=-\frac{1}{8 \pi G}\left(\partial_{\mu} \Phi\right)^{2}+T \Phi \tag{E.394} \end{equation*}(E.394)L=18πG(μΦ)2+TΦ
(43.2) Using eqn 43.51 the equation in question can be rearranged and reduced to read
(E.395) R α ν μ ; β β R α μ ; ν + R α ν ; μ = 0 (E.395) R α ν μ ; β β R α μ ; ν + R α ν ; μ = 0 {:(E.395)R_(alpha nu mu;beta)^(beta)-R_(alpha mu;nu)+R_(alpha nu;mu)=0:}\begin{equation*} R_{\alpha \nu \mu ; \beta}^{\beta}-R_{\alpha \mu ; \nu}+R_{\alpha \nu ; \mu}=0 \tag{E.395} \end{equation*}(E.395)Rανμ;ββRαμ;ν+Rαν;μ=0
This can be recreated from the contraction of the Bianchi identity R α [ β ν ; μ ] β = 0 R α [ β ν ; μ ] β = 0 R_(alpha[beta nu;mu])^(beta)=0R_{\alpha[\beta \nu ; \mu]}^{\beta}=0Rα[βν;μ]β=0.
(43.3) Taking the dual with respect to the vector part of the ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) tensor (and raising an index) will result in a ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor T = ( T ) ν α β γ ( e ν ω α ω β ω γ ) T = ( T ) ν α β γ e ν ω α ω β ω γ ***T=(***T)^(nu)_(alpha beta gamma)(e_(nu)oxomega^(alpha)^^omega^(beta)^^omega^(gamma))\star \boldsymbol{T}=(\star T)^{\nu}{ }_{\alpha \beta \gamma}\left(\boldsymbol{e}_{\nu} \otimes \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} \wedge \boldsymbol{\omega}^{\gamma}\right)T=(T)ναβγ(eνωαωβωγ) with components
(E.396) ( T ) ν α β γ = 1 3 ! g ε μ α β γ T μ ν (E.396) ( T ) ν α β γ = 1 3 ! g ε μ α β γ T μ ν {:(E.396)(***T)^(nu)_(alpha beta gamma)=(1)/(3!)sqrt(-g)epsi_(mu alpha beta gamma)T^(mu nu):}\begin{equation*} (\star T)^{\nu}{ }_{\alpha \beta \gamma}=\frac{1}{3!} \sqrt{-g} \varepsilon_{\mu \alpha \beta \gamma} T^{\mu \nu} \tag{E.396} \end{equation*}(E.396)(T)ναβγ=13!gεμαβγTμν
The expressions for the components of G G ***G\star \boldsymbol{G}G are similar.
(43.4) Apply the operator d d d\boldsymbol{d}d to T μ ν ( e μ d Σ ν ) T μ ν e μ d Σ ν T^(mu nu)(e_(mu)oxdSigma_(nu))T^{\mu \nu}\left(\boldsymbol{e}_{\mu} \otimes \mathrm{d} \boldsymbol{\Sigma}_{\nu}\right)Tμν(eμdΣν). We obtain
[ d T μ ν e μ g + T μ ν d e μ g + e μ T μ ν d ( g ) ] ε ν | α β γ | d x α d x β d x γ . d T μ ν e μ g + T μ ν d e μ g + e μ T μ ν d ( g ) ε ν | α β γ | d x α d x β d x γ . {:[[dT^(mu nu)e_(mu)sqrt(-g)+T^(mu nu)de_(mu)sqrt(-g)+e_(mu)T^(mu nu)d(sqrt(-g))]],[ oxepsi_(nu|alpha beta gamma|)dx^(alpha)^^dx^(beta)^^dx^(gamma).]:}\begin{aligned} & {\left[\boldsymbol{d} T^{\mu \nu} \boldsymbol{e}_{\mu} \sqrt{-g}+T^{\mu \nu} \boldsymbol{d} \boldsymbol{e}_{\mu} \sqrt{-g}+\boldsymbol{e}_{\mu} T^{\mu \nu} \boldsymbol{d}(\sqrt{-g})\right]} \\ & \otimes \varepsilon_{\nu|\alpha \beta \gamma|} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \boldsymbol{d} x^{\gamma} . \end{aligned}[dTμνeμg+Tμνdeμg+eμTμνd(g)]εν|αβγ|dxαdxβdxγ.
The first of the exterior derivatives yields (with some changes in contracted index names in the coefficients)
g e μ d T μ ν = g T μ λ x λ e μ d x ν , (E.398) g e μ d T μ ν = g T μ λ x λ e μ d x ν ,  (E.398)  sqrt(-g)e_(mu)dT^(mu nu)=sqrt(-g)(delT^(mu lambda))/(delx^(lambda))e_(mu)ox dx^(nu),quad" (E.398) "\sqrt{-g} \boldsymbol{e}_{\mu} \boldsymbol{d} T^{\mu \nu}=\sqrt{-g} \frac{\partial T^{\mu \lambda}}{\partial x^{\lambda}} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}, \quad \text { (E.398) }geμdTμν=gTμλxλeμdxν, (E.398) 
the second is
g T μ ν d e μ = g Γ λ σ μ T σ λ e μ d x ν , g T μ ν d e μ = g Γ λ σ μ T σ λ e μ d x ν , sqrt(-g)T^(mu nu)de_(mu)=sqrt(-g)Gamma_(lambda sigma)^(mu)T^(sigma lambda)e_(mu)ox dx^(nu),quad\sqrt{-g} T^{\mu \nu} \boldsymbol{d} e_{\mu}=\sqrt{-g} \Gamma_{\lambda \sigma}^{\mu} T^{\sigma \lambda} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}, \quadgTμνdeμ=gΓλσμTσλeμdxν, (E.399) and the final one is
e μ T μ ν d g = g Γ λ σ λ T σ μ e μ d x ν e μ T μ ν d g = g Γ λ σ λ T σ μ e μ d x ν e_(mu)T^(mu nu)dsqrt(-g)=sqrt(-g)Gamma_(lambda sigma)^(lambda)T^(sigma mu)e_(mu)ox dx^(nu)\boldsymbol{e}_{\mu} T^{\mu \nu} \boldsymbol{d} \sqrt{-g}=\sqrt{-g} \Gamma_{\lambda \sigma}^{\lambda} T^{\sigma \mu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}eμTμνdg=gΓλσλTσμeμdxν
So we obtain T μ λ ; λ e μ d x ν T μ λ ; λ e μ d x ν T^(mu lambda)_(;lambda)e_(mu)ox dx^(nu)T^{\mu \lambda}{ }_{; \lambda} \boldsymbol{e}_{\mu} \otimes \boldsymbol{d} x^{\nu}Tμλ;λeμdxν. Consider the wedge product: the only parts that survive will have ν ν nu\nuν distinct from α , β α , β alpha,beta\alpha, \betaα,β and γ γ gamma\gammaγ. We therefore obtain
d T = e μ T ; ν μ ν g d x 0 d x 1 d x 2 d x 3 d T = e μ T ; ν μ ν g d x 0 d x 1 d x 2 d x 3 d***T=e_(mu)T_(;nu)^(mu nu)sqrt(-g)dx^(0)^^dx^(1)^^dx^(2)^^dx^(3)\boldsymbol{d} \star \boldsymbol{T}=\boldsymbol{e}_{\mu} T_{; \nu}^{\mu \nu} \sqrt{-g} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}dT=eμT;νμνgdx0dx1dx2dx3
The components are all zero by the usual conservation law T μ ν ; ν = 0 T μ ν ; ν = 0 T^(mu nu)_(;nu)=0T^{\mu \nu}{ }_{; \nu}=0Tμν;ν=0.
(43.5) We run the argument from the chapter. The conservation law, written as a
(E.402) V d T = 0 (E.402) V d T = 0 {:(E.402)int_(V)d***T=0:}\begin{equation*} \int_{\mathcal{V}} d \star T=0 \tag{E.402} \end{equation*}(E.402)VdT=0
Using the Einstein equation
V d G = 0 , V d G = 0 , int_(V)d***G=0,\int_{\mathcal{V}} d \star G=0,VdG=0,
and, using Stokes' theorem,
V G = 0 V G = 0 int_(delV)***G=0\int_{\partial \mathcal{V}} \star \boldsymbol{G}=0VG=0
(44.1) (a) Using the technique in the chapter, the effect of going round the loop is
(E.405) ψ A 1 = ( 1 i g Δ S μ ν G μ ν ) ψ A 0 (E.405) ψ A 1 = 1 i g Δ S μ ν G μ ν ψ A 0 {:(E.405)psi_(A1)=(1-ig DeltaS^(mu nu)G_(mu nu))psi_(A0):}\begin{equation*} \psi_{\mathcal{A} 1}=\left(1-\mathrm{i} g \Delta S^{\mu \nu} G_{\mu \nu}\right) \psi_{\mathcal{A} 0} \tag{E.405} \end{equation*}(E.405)ψA1=(1igΔSμνGμν)ψA0
where Δ S μ ν = δ x μ Δ x ν Δ S μ ν = δ x μ Δ x ν DeltaS^(mu nu)=deltax^(mu)Deltax^(nu)\Delta S^{\mu \nu}=\delta x^{\mu} \Delta x^{\nu}ΔSμν=δxμΔxν is the area A B C D A B C D ABCD\mathcal{A B C D}ABCD. (Here the subscripts give the point where the field is evaluated along with a number denoting which step we're doing in the computation.)
(b) The result here is simply
(E.406) ψ P 2 = ( 1 + d x ρ D ρ ) ψ A 1 (E.406) ψ P 2 = 1 + d x ρ D ρ ψ A 1 {:(E.406)psi_(P2)=(1+dx^(rho)D_(rho))psi_(A1):}\begin{equation*} \psi_{\mathcal{P} 2}=\left(1+\mathrm{d} x^{\rho} D_{\rho}\right) \psi_{\mathcal{A} 1} \tag{E.406} \end{equation*}(E.406)ψP2=(1+dxρDρ)ψA1
(c) Similarly to part (a) we have
(E.407) ψ P 3 = ( 1 + i g Δ S μ ν G μ ν ) ψ P 2 (E.407) ψ P 3 = 1 + i g Δ S μ ν G μ ν ψ P 2 {:(E.407)psi_(P3)=(1+ig DeltaS^(mu nu)G_(mu nu))psi_(P2):}\begin{equation*} \psi_{\mathcal{P} 3}=\left(1+\mathrm{i} g \Delta S^{\mu \nu} G_{\mu \nu}\right) \psi_{\mathcal{P} 2} \tag{E.407} \end{equation*}(E.407)ψP3=(1+igΔSμνGμν)ψP2
(d) Similarly to part (b) we have
ψ A 4 = ( 1 d x σ D σ ) ψ P 3 ψ A 4 = 1 d x σ D σ ψ P 3 psi_(A4)=(1-dx^(sigma)D_(sigma))psi_(P3)\psi_{\mathcal{A} 4}=\left(1-\mathrm{d} x^{\sigma} D_{\sigma}\right) \psi_{\mathcal{P} 3}ψA4=(1dxσDσ)ψP3
(e) We find
ψ A 4 = ( 1 d x σ D σ ) ( 1 + i g Δ S μ ν G μ ν ) × ( 1 + d x ρ D ρ ) ( 1 i g Δ S μ ν G μ ν ) ψ A 0 = ( 1 i g Δ V ρ μ ν [ D ρ , G μ ν ] ) ψ A 0 , (E.409) ψ A 4 = 1 d x σ D σ 1 + i g Δ S μ ν G μ ν × 1 + d x ρ D ρ 1 i g Δ S μ ν G μ ν ψ A 0 = 1 i g Δ V ρ μ ν D ρ , G μ ν ψ A 0 ,  (E.409)  {:[psi_(A4)=(1-dx^(sigma)D_(sigma))(1+ig DeltaS^(mu nu)G_(mu nu))],[ xx(1+dx^(rho)D_(rho))(1-ig DeltaS^(mu nu)G_(mu nu))psi_(A0)],[=(1-ig DeltaV^(rho mu nu)[D_(rho),G_(mu nu)])psi_(A0)","quad" (E.409) "]:}\begin{aligned} \psi_{\mathcal{A} 4}= & \left(1-\mathrm{d} x^{\sigma} D_{\sigma}\right)\left(1+\mathrm{i} g \Delta S^{\mu \nu} G_{\mu \nu}\right) \\ & \times\left(1+\mathrm{d} x^{\rho} D_{\rho}\right)\left(1-\mathrm{i} g \Delta S^{\mu \nu} G_{\mu \nu}\right) \psi_{\mathcal{A} 0} \\ = & \left(1-\mathrm{i} g \Delta V^{\rho \mu \nu}\left[D_{\rho}, G_{\mu \nu}\right]\right) \psi_{\mathcal{A} 0}, \quad \text { (E.409) } \end{aligned}ψA4=(1dxσDσ)(1+igΔSμνGμν)×(1+dxρDρ)(1igΔSμνGμν)ψA0=(1igΔVρμν[Dρ,Gμν])ψA0, (E.409) 
where V ρ μ ν = d x ρ δ x μ Δ x ν V ρ μ ν = d x ρ δ x μ Δ x ν V^(rho mu nu)=dx^(rho)deltax^(mu)Deltax^(nu)V^{\rho \mu \nu}=\mathrm{d} x^{\rho} \delta x^{\mu} \Delta x^{\nu}Vρμν=dxρδxμΔxν is the volume of the cube. Since the differential operator acts on G μ ν G μ ν G_(mu nu)G_{\mu \nu}Gμν and also ψ A 0 ψ A 0 psi_(A0)\psi_{\mathcal{A} 0}ψA0, we may write, approximately, that
(E.410) ψ A 4 = ( 1 i g Δ V ρ μ ν D ρ G μ ν ) ψ A 0 (E.410) ψ A 4 = 1 i g Δ V ρ μ ν D ρ G μ ν ψ A 0 {:(E.410)psi_(A4)=(1-ig DeltaV^(rho mu nu)D_(rho)G_(mu nu))psi_(A0):}\begin{equation*} \psi_{\mathcal{A} 4}=\left(1-\mathrm{i} g \Delta V^{\rho \mu \nu} D_{\rho} G_{\mu \nu}\right) \psi_{\mathcal{A} 0} \tag{E.410} \end{equation*}(E.410)ψA4=(1igΔVρμνDρGμν)ψA0
(g) The argument in Chapter 43 tells us that traversing the path makes zero contribution to the field, so the bracket in eqn 44.26 can be set to zero. Substitute for G μ ν G μ ν G_(mu nu)G_{\mu \nu}Gμν and confirm that the expressions are identical.
(44.2) The symmetrized energy-momentum tensor has components
(E.411) T μ ν = ( ψ , μ ψ , ν + ψ , μ ψ , ν ) + g μ ν L (E.411) T μ ν = ψ , μ ψ , ν + ψ , μ ψ , ν + g μ ν L {:(E.411)T_(mu nu)=(psi_(,mu)^(†)psi_(,nu)+psi_(,mu)psi_(,nu)^(†))+g_(mu nu)L:}\begin{equation*} T_{\mu \nu}=\left(\psi_{, \mu}^{\dagger} \psi_{, \nu}+\psi_{, \mu} \psi_{, \nu}^{\dagger}\right)+g_{\mu \nu} \mathcal{L} \tag{E.411} \end{equation*}(E.411)Tμν=(ψ,μψ,ν+ψ,μψ,ν)+gμνL
(45.3) Use the fact that T μ ν = ( ρ + p ) u μ u ν + g μ ν p T μ ν = ( ρ + p ) u μ u ν + g μ ν p T_(mu nu)=(rho+p)u_(mu)u_(nu)+g_(mu nu)pT_{\mu \nu}=(\rho+p) u_{\mu} u_{\nu}+g_{\mu \nu} pTμν=(ρ+p)uμuν+gμνp to find the following:
(a) ρ g = 2 ρ + ( ρ + 3 p ) ρ g = 2 ρ + ( ρ + 3 p ) rho_(g)=2rho+(-rho+3p)\rho_{\mathrm{g}}=2 \rho+(-\rho+3 p)ρg=2ρ+(ρ+3p).
(b) Π i = ( ρ + p ) u i u 0 Π i = ( ρ + p ) u i u 0 Pi_(i)=-(rho+p)u_(i)u_(0)\Pi_{i}=-(\rho+p) u_{i} u_{0}Πi=(ρ+p)uiu0 where, at low velocity, we have u μ = ( 1 , u ) u μ = ( 1 , u ) u^(mu)=(1, vec(u))u^{\mu}=(1, \vec{u})uμ=(1,u) and u μ = ( 1 , u ) u μ = ( 1 , u ) u_(mu)=(-1, vec(u))u_{\mu}=(-1, \vec{u})uμ=(1,u).
(c) ρ c = 2 p ( ρ + 3 p ) ρ c = 2 p ( ρ + 3 p ) rho_(c)=2p-(-rho+3p)\rho_{\mathrm{c}}=2 p-(-\rho+3 p)ρc=2p(ρ+3p).
(45.4) The key here is that we only retain terms up to order u i u i vec(u)^(i)\vec{u}^{i}ui, ignoring those at order u 2 u 2 u^(2)u^{2}u2. In the weak-field, lowvelocity limit we have a geodesic equation
2 x μ t 2 + Γ α β μ u α u β = 0 , 2 x μ t 2 + Γ α β μ u α u β = 0 , (del^(2)x^(mu))/(delt^(2))+Gamma_(alpha beta)^(mu)u^(alpha)u^(beta)=0,\frac{\partial^{2} x^{\mu}}{\partial t^{2}}+\Gamma_{\alpha \beta}^{\mu} u^{\alpha} u^{\beta}=0,2xμt2+Γαβμuαuβ=0,
(E.412)
with u α = ( 1 , u ) = ( 1 , u i ) u α = ( 1 , u ) = 1 , u i u^(alpha)=(1, vec(u))=(1,u^(i))u^{\alpha}=(1, \vec{u})=\left(1, u^{i}\right)uα=(1,u)=(1,ui) and
(E.413) Γ α β μ = 1 2 η μ ν ( h α ν , β + h β ν , α h α β , ν ) (E.413) Γ α β μ = 1 2 η μ ν h α ν , β + h β ν , α h α β , ν {:(E.413)Gamma_(alpha beta)^(mu)=(1)/(2)eta^(mu nu)(h_(alpha nu,beta)+h_(beta nu,alpha)-h_(alpha beta,nu)):}\begin{equation*} \Gamma_{\alpha \beta}^{\mu}=\frac{1}{2} \eta^{\mu \nu}\left(h_{\alpha \nu, \beta}+h_{\beta \nu, \alpha}-h_{\alpha \beta, \nu}\right) \tag{E.413} \end{equation*}(E.413)Γαβμ=12ημν(hαν,β+hβν,αhαβ,ν)
We find
(E.414) Γ i t t = 1 2 η i j h t t , j (E.414) Γ i t t = 1 2 η i j h t t , j {:(E.414)Gamma^(i)_(tt)=-(1)/(2)eta^(ij)h_(tt,j):}\begin{equation*} \Gamma^{i}{ }_{t t}=-\frac{1}{2} \eta^{i j} h_{t t, j} \tag{E.414} \end{equation*}(E.414)Γitt=12ηijhtt,j
and
(E.415) Γ t k i = 1 2 η i j ( h t j , k h t k , j ) (E.415) Γ t k i = 1 2 η i j h t j , k h t k , j {:(E.415)Gamma_(tk)^(i)=(1)/(2)eta^(ij)(h_(tj,k)-h_(tk,j)):}\begin{equation*} \Gamma_{t k}^{i}=\frac{1}{2} \eta^{i j}\left(h_{t j, k}-h_{t k, j}\right) \tag{E.415} \end{equation*}(E.415)Γtki=12ηij(htj,khtk,j)
geodesic equation and retaining terms up to order u i u i u^(i)u^{i}ui geodesic equation and retaining terms up to order u i u i u^(i)u^{i}ui (b) From the answer
(E.416) 2 h t t = 16 π ( T t t 1 2 η t t T ) (E.416) 2 h t t = 16 π T t t 1 2 η t t T {:(E.416)grad^(2)h_(tt)=-16 pi(T_(tt)-(1)/(2)*eta_(tt)T):}\begin{equation*} \nabla^{2} h_{t t}=-16 \pi\left(T_{t t}-\frac{1}{2} \cdot \eta_{t t} T\right) \tag{E.416} \end{equation*}(E.416)2htt=16π(Ttt12ηttT)
we obtain Poisson's equation, whose solution is the expression given.
(45.5) (a) Start with
(E.417) 2 h μ ν = 8 π ( T μ ν 1 2 η μ ν T ) Λ η μ ν (E.417) 2 h μ ν = 8 π T μ ν 1 2 η μ ν T Λ η μ ν {:(E.417)-del^(2)h_(mu nu)=8pi(T_(mu nu)-(1)/(2)*eta_(mu nu)T)-Lambdaeta_(mu nu):}\begin{equation*} -\partial^{2} h_{\mu \nu}=8 \pi\left(T_{\mu \nu}-\frac{1}{2} \cdot \eta_{\mu \nu} T\right)-\Lambda \eta_{\mu \nu} \tag{E.417} \end{equation*}(E.417)2hμν=8π(Tμν12ημνT)Λημν
If, in the absence of matter, we were to take the effective energy density to be T μ ν = Λ 8 π η μ ν T μ ν = Λ 8 π η μ ν T_(mu nu)^(')=-(Lambda)/(8pi)eta_(mu nu)T_{\mu \nu}^{\prime}=-\frac{\Lambda}{8 \pi} \eta_{\mu \nu}Tμν=Λ8πημν, then we have a trace T = Λ / 2 π T = Λ / 2 π T^(')=-Lambda//2piT^{\prime}=-\Lambda / 2 \piT=Λ/2π and find an energy density ρ g = 2 T t t η t t T = Λ / 4 π ρ g = 2 T t t η t t T = Λ / 4 π rho_(g)=2T_(tt)^(')-eta_(tt)T^(')=-Lambda//4pi\rho_{\mathrm{g}}=2 T_{t t}^{\prime}-\eta_{t t} T^{\prime}=-\Lambda / 4 \piρg=2TttηttT=Λ/4π. This is negative and so the force is repulsive.
(b) We find a field equation 2 h t t = 2 Λ 2 h t t = 2 Λ vec(grad)^(2)h_(tt)=2Lambda\vec{\nabla}^{2} h_{t t}=2 \Lambda2htt=2Λ and h t t = 2 Φ h t t = 2 Φ h_(tt)=-2Phih_{t t}=-2 \Phihtt=2Φ, giving 2 Φ = Λ 2 Φ = Λ vec(grad)^(2)Phi=-Lambda\vec{\nabla}^{2} \Phi=-\Lambda2Φ=Λ. The function Φ = α r 2 Φ = α r 2 Phi=alphar^(2)\Phi=\alpha r^{2}Φ=αr2 solves the resulting equation
(E.418) 1 r 2 r ( r 2 Φ r ) = Λ (E.418) 1 r 2 r r 2 Φ r = Λ {:(E.418)(1)/(r^(2))(del)/(del r)*(r^(2)(del Phi)/(del r))=-Lambda:}\begin{equation*} \frac{1}{r^{2}} \frac{\partial}{\partial r} \cdot\left(r^{2} \frac{\partial \Phi}{\partial r}\right)=-\Lambda \tag{E.418} \end{equation*}(E.418)1r2r(r2Φr)=Λ
for α = Λ / 6 α = Λ / 6 alpha=-Lambda//6\alpha=-\Lambda / 6α=Λ/6.
(45.6) (b) Using the usual formula for Γ μ ν σ Γ μ ν σ Gamma_(mu nu sigma)\Gamma_{\mu \nu \sigma}Γμνσ in terms of the components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν, we find
Γ 0 j k = 1 2 ( ζ j , k + ζ k , j ) Γ i 0 k = 1 2 ( ζ j , k ζ k , j ) Γ i 00 = ϕ , i Γ 0 i 0 = ϕ , i (E.419) Γ i j k == ϕ , k δ i j ϕ , j δ i k ϕ , i δ j k Γ 0 j k = 1 2 ζ j , k + ζ k , j Γ i 0 k = 1 2 ζ j , k ζ k , j Γ i 00 = ϕ , i Γ 0 i 0 = ϕ , i (E.419) Γ i j k == ϕ , k δ i j ϕ , j δ i k ϕ , i δ j k {:[Gamma_(0jk)=(1)/(2)(zeta_(j,k)+zeta_(k,j))],[Gamma_(i0k)=(1)/(2)*(zeta_(j,k)-zeta_(k,j))],[Gamma_(i 00)=phi_(,i)],[Gamma_(0i0)=-phi_(,i)],[(E.419)Gamma_(ijk)==-phi_(,k)delta_(ij)-phi_(,j)delta_(ik)-phi_(,i)delta_(jk)]:}\begin{align*} \Gamma_{0 j k} & =\frac{1}{2}\left(\zeta_{j, k}+\zeta_{k, j}\right) \\ \Gamma_{i 0 k} & =\frac{1}{2} \cdot\left(\zeta_{j, k}-\zeta_{k, j}\right) \\ \Gamma_{i 00} & =\phi_{, i} \\ \Gamma_{0 i 0} & =-\phi_{, i} \\ \Gamma_{i j k} & ==-\phi_{, k} \delta_{i j}-\phi_{, j} \delta_{i k}-\phi_{, i} \delta_{j k} \tag{E.419} \end{align*}Γ0jk=12(ζj,k+ζk,j)Γi0k=12(ζj,kζk,j)Γi00=ϕ,iΓ0i0=ϕ,i(E.419)Γijk==ϕ,kδijϕ,jδikϕ,iδjk
To the order at which we're working, raising the index can be done with the Minkowski tensor. (Alternatively, use g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν and drop higher order terms.)
(45.7) (a) Use S μ u μ = 0 S μ u μ = 0 S_(mu)u^(mu)=0S_{\mu} u^{\mu}=0Sμuμ=0 to say
S 0 d t d τ = S i d x i d τ S 0 = S i d x i d τ d τ d t = S i d x i d t S 0 d t d τ = S i d x i d τ S 0 = S i d x i d τ d τ d t = S i d x i d t {:[S_(0)((d)t)/((d)tau)=-S_(i)((d)x^(i))/((d)tau)],[S_(0)=-S_(i)((d)x^(i))/((d)tau)*((d)tau)/((d)t)=-S_(i)((d)x^(i))/((d)t)]:}\begin{aligned} S_{0} \frac{\mathrm{~d} t}{\mathrm{~d} \tau} & =-S_{i} \frac{\mathrm{~d} x^{i}}{\mathrm{~d} \tau} \\ S_{0} & =-S_{i} \frac{\mathrm{~d} x^{i}}{\mathrm{~d} \tau} \cdot \frac{\mathrm{~d} \tau}{\mathrm{~d} t}=-S_{i} \frac{\mathrm{~d} x^{i}}{\mathrm{~d} t} \end{aligned}S0 dt dτ=Si dxi dτS0=Si dxi dτ dτ dt=Si dxi dt
(E.420)
(b) For parallel transport of a 1-form, we have
( S μ x ν Γ ν μ σ S σ ) u ν = 0 S μ x ν Γ ν μ σ S σ u ν = 0 ((delS_(mu))/(delx^(nu))-Gamma_(nu mu)^(sigma)S_(sigma))u^(nu)=0\left(\frac{\partial S_{\mu}}{\partial x^{\nu}}-\Gamma_{\nu \mu}^{\sigma} S_{\sigma}\right) u^{\nu}=0(SμxνΓνμσSσ)uν=0
Rearranging
S μ x ν d x ν d τ = Γ σ ν μ S σ d x ν d τ d S μ d τ = Γ σ ν μ S σ d x ν d τ d S μ d t = Γ σ ν μ S σ d x ν d τ d τ d t (E.422) = Γ σ ν μ S σ d x ν d t S μ x ν d x ν d τ = Γ σ ν μ S σ d x ν d τ d S μ d τ = Γ σ ν μ S σ d x ν d τ d S μ d t = Γ σ ν μ S σ d x ν d τ d τ d t (E.422) = Γ σ ν μ S σ d x ν d t {:[(delS_(mu))/(delx^(nu))(dx^(nu))/(dtau)=Gamma^(sigma)_(nu mu)S_(sigma)(dx^(nu))/(dtau)],[((d)S_(mu))/(dtau)=Gamma^(sigma)_(nu mu)S_(sigma)(dx^(nu))/(dtau)],[((d)S_(mu))/(dt)=Gamma^(sigma)_(nu mu)S_(sigma)(dx^(nu))/(dtau)*((d)tau)/((d)t)],[(E.422)=Gamma^(sigma)_(nu mu)S_(sigma)(dx^(nu))/(dt)]:}\begin{align*} \frac{\partial S_{\mu}}{\partial x^{\nu}} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} & =\Gamma^{\sigma}{ }_{\nu \mu} S_{\sigma} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} \\ \frac{\mathrm{~d} S_{\mu}}{\mathrm{d} \tau} & =\Gamma^{\sigma}{ }_{\nu \mu} S_{\sigma} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} \\ \frac{\mathrm{~d} S_{\mu}}{\mathrm{d} t} & =\Gamma^{\sigma}{ }_{\nu \mu} S_{\sigma} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau} \cdot \frac{\mathrm{~d} \tau}{\mathrm{~d} t} \\ & =\Gamma^{\sigma}{ }_{\nu \mu} S_{\sigma} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} t} \tag{E.422} \end{align*}Sμxνdxνdτ=ΓσνμSσdxνdτ dSμdτ=ΓσνμSσdxνdτ dSμdt=ΓσνμSσdxνdτ dτ dt(E.422)=ΓσνμSσdxνdt
(45.8) Parallel transport preserves the magnitude of a vector, so we can say that g μ ν S μ S ν g μ ν S μ S ν g^(mu nu)S_(mu)S_(nu)g^{\mu \nu} S_{\mu} S_{\nu}gμνSμSν is constant, which means
g 00 ( S 0 ) 2 + g i j S i S j = const. g 00 S 0 2 + g i j S i S j =  const.  g^(00)(S_(0))^(2)+g^(ij)S_(i)S_(j)=" const. "g^{00}\left(S_{0}\right)^{2}+g^{i j} S_{i} S_{j}=\text { const. }g00(S0)2+gijSiSj= const. 
Then use S 0 = ( d x i / d t ) S i S 0 = d x i / d t S i S_(0)=-(dx^(i)//dt)S_(i)S_{0}=-\left(\mathrm{d} x^{i} / \mathrm{d} t\right) S_{i}S0=(dxi/dt)Si and v i = d x i / d t v i = d x i / d t v^(i)=dx^(i)//dtv^{i}=\mathrm{d} x^{i} / \mathrm{d} tvi=dxi/dt and ignore the non-diagonal term with g i j g i j g^(ij)g^{i j}gij, since this is higher order. Note also that in writing the 3 -vector, indices can der. Note also that in writing the 3 -vector, indices can
be raised and lowered with the Minkowski metric tenbe raised and lo
sor, so v i = v i v i = v i v_(i)=v^(i)v_{i}=v^{i}vi=vi.
(45.9) (a) Dot the equation for S S vec(S)\vec{S}S with itself.
(b) Compute v S v S vec(v)* vec(S)\vec{v} \cdot \vec{S}vS.
(c) Combine (a) and (b) with eqn 45.60
(e) Taking a derivative, we have
d Σ d t = d S d t + ( ϕ v ) S + 1 2 ( v S ) ϕ (E.424) + 1 2 ( ϕ S ) v d Σ d t = d S d t + ( ϕ v ) S + 1 2 ( v S ) ϕ (E.424) + 1 2 ( ϕ S ) v {:[(d( vec(Sigma)))/((d)t)=(d( vec(S)))/((d)t)+( vec(grad)phi* vec(v)) vec(S)+(1)/(2)( vec(v)* vec(S)) vec(grad)phi],[(E.424)+(1)/(2)*( vec(grad)phi* vec(S)) vec(v)]:}\begin{align*} \frac{\mathrm{d} \vec{\Sigma}}{\mathrm{~d} t}= & \frac{\mathrm{d} \vec{S}}{\mathrm{~d} t}+(\vec{\nabla} \phi \cdot \vec{v}) \vec{S}+\frac{1}{2}(\vec{v} \cdot \vec{S}) \vec{\nabla} \phi \\ & +\frac{1}{2} \cdot(\vec{\nabla} \phi \cdot \vec{S}) \vec{v} \tag{E.424} \end{align*}dΣ dt=dS dt+(ϕv)S+12(vS)ϕ(E.424)+12(ϕS)v
In vector notation,
d S d t = 2 ( v S ) ϕ + ( ϕ S ) v ( ϕ v ) S + 1 2 [ S × ( × ζ ) ] . (E.425) Combine these to find d S d t = 2 ( v S ) ϕ + ( ϕ S ) v ( ϕ v ) S + 1 2 [ S × ( × ζ ) ] . (E.425)  Combine these to find  {:[qquad{:[(d( vec(S)))/((d)t)=-2( vec(v)* vec(S)) vec(grad)phi+( vec(grad)phi* vec(S)) vec(v)-( vec(grad)phi* vec(v)) vec(S)],[+(1)/(2)[ vec(S)xx( vec(grad)xx vec(zeta))].]:}],[(E.425)" Combine these to find "]:}\begin{align*} & \qquad \begin{aligned} \frac{\mathrm{d} \vec{S}}{\mathrm{~d} t}= & -2(\vec{v} \cdot \vec{S}) \vec{\nabla} \phi+(\vec{\nabla} \phi \cdot \vec{S}) \vec{v}-(\vec{\nabla} \phi \cdot \vec{v}) \vec{S} \\ & +\frac{1}{2}[\vec{S} \times(\vec{\nabla} \times \vec{\zeta})] . \end{aligned} \\ & \text { Combine these to find } \tag{E.425} \end{align*}dS dt=2(vS)ϕ+(ϕS)v(ϕv)S+12[S×(×ζ)].(E.425) Combine these to find 
d Σ d t = 1 2 [ S × ( × ζ ) ] (E.426) 3 2 [ ( v S ) ϕ ( ϕ S ) v ) ] d Σ d t = 1 2 [ S × ( × ζ ) ] (E.426) 3 2 [ ( v S ) ϕ ( ϕ S ) v ) {:[(d( vec(Sigma)))/((d)t)=(1)/(2)[ vec(S)xx( vec(grad)xx vec(zeta))]],[(E.426){:-(3)/(2)[(( vec(v))*( vec(S)))( vec(grad))phi-(( vec(grad))phi*( vec(S)))( vec(v)))]]:}\begin{align*} \frac{\mathrm{d} \vec{\Sigma}}{\mathrm{~d} t}= & \frac{1}{2}[\vec{S} \times(\vec{\nabla} \times \vec{\zeta})] \\ & \left.-\frac{3}{2}[(\vec{v} \cdot \vec{S}) \vec{\nabla} \phi-(\vec{\nabla} \phi \cdot \vec{S}) \vec{v})\right] \tag{E.426} \end{align*}dΣ dt=12[S×(×ζ)](E.426)32[(vS)ϕ(ϕS)v)]
Using the rule for a double vector product, we can spot that the final two terms are equal to 3 2 [ S × ( v × ϕ ] 3 2 [ S × ( v × ϕ ] (3)/(2)[ vec(S)xx( vec(v)xx vec(grad)phi]\frac{3}{2}[\vec{S} \times(\vec{v} \times \vec{\nabla} \phi]32[S×(v×ϕ] and the answer follows.
(f) First, ϕ = M r / r 3 ϕ = M r / r 3 vec(grad)phi=M vec(r)//r^(3)\vec{\nabla} \phi=M \vec{r} / r^{3}ϕ=Mr/r3. Then use components to say
( × ζ ) a = b c ε a b c b ζ c = b c k m ε a b c x b 1 r 3 ε c k m x k J m = b c k m ε c b a ε c k m x b ( x k r 3 ) J m (E.427) = b c k m ε c b a ε c k m ( r 3 x k x b 3 x k x b r ) J m r 6 . ( × ζ ) a = b c ε a b c b ζ c = b c k m ε a b c x b 1 r 3 ε c k m x k J m = b c k m ε c b a ε c k m x b x k r 3 J m (E.427) = b c k m ε c b a ε c k m r 3 x k x b 3 x k x b r J m r 6 . {:[( vec(grad)xx zeta)_(a)=sum_(bc)epsi_(abc)grad_(b)zeta_(c)],[=sum_(bckm)epsi_(abc)(del)/(delx^(b))(1)/(r^(3))epsi_(ckm)x^(k)J^(m)],[=-sum_(bckm)epsi_(cba)epsi_(ckm)(del)/(delx^(b))*((x^(k))/(r^(3)))J^(m)],[(E.427)=-sum_(bckm)epsi_(cba)epsi_(ckm)(r^(3)(delx^(k))/(delx^(b))-3x^(k)x^(b)r)(J^(m))/(r^(6)).]:}\begin{align*} (\vec{\nabla} \times \zeta)_{a} & =\sum_{b c} \varepsilon_{a b c} \nabla_{b} \zeta_{c} \\ & =\sum_{b c k m} \varepsilon_{a b c} \frac{\partial}{\partial x^{b}} \frac{1}{r^{3}} \varepsilon_{c k m} x^{k} J^{m} \\ & =-\sum_{b c k m} \varepsilon_{c b a} \varepsilon_{c k m} \frac{\partial}{\partial x^{b}} \cdot\left(\frac{x^{k}}{r^{3}}\right) J^{m} \\ & =-\sum_{b c k m} \varepsilon_{c b a} \varepsilon_{c k m}\left(r^{3} \frac{\partial x^{k}}{\partial x^{b}}-3 x^{k} x^{b} r\right) \frac{J^{m}}{r^{6}} . \tag{E.427} \end{align*}(×ζ)a=bcεabcbζc=bckmεabcxb1r3εckmxkJm=bckmεcbaεckmxb(xkr3)Jm(E.427)=bckmεcbaεckm(r3xkxb3xkxbr)Jmr6.
Then use the rule that c ε c b a ε c k m = δ k b δ m a δ m b δ k a c ε c b a ε c k m = δ k b δ m a δ m b δ k a sum_(c)epsi_(cba)epsi_(ckm)=delta_(k)^(b)delta_(m)^(a)-delta_(m)^(b)delta_(k)^(a)\sum_{c} \varepsilon_{c b a} \varepsilon_{c k m}=\delta_{k}^{b} \delta_{m}^{a}-\delta_{m}^{b} \delta_{k}^{a}cεcbaεckm=δkbδmaδmbδka to (47.4) (a) The centre of mass is defined by m 1 a 1 = m 2 a 2 m 1 a 1 = m 2 a 2 m_(1)a_(1)=m_(2)a_(2)m_{1} a_{1}=m_{2} a_{2}m1a1=m2a2 obtain
( × ζ ) a = b [ r 3 x b x b 3 ( x b ) 2 r ] J a r 6 + b [ r 3 x a x b 3 x a x b r ] J b r 6 , (E.428) ( × ζ ) a = b r 3 x b x b 3 x b 2 r J a r 6 + b r 3 x a x b 3 x a x b r J b r 6 ,  (E.428)  {:[( vec(grad)xx zeta)_(a)=-sum_(b)[r^(3)(delx^(b))/(delx^(b))-3(x^(b))^(2)r](J^(a))/(r^(6))],[+sum_(b)[r^(3)(delx^(a))/(delx^(b))-3x^(a)x^(b)r](J^(b))/(r^(6))","quad" (E.428) "]:}\begin{aligned} (\vec{\nabla} \times \zeta)_{a}= & -\sum_{b}\left[r^{3} \frac{\partial x^{b}}{\partial x^{b}}-3\left(x^{b}\right)^{2} r\right] \frac{J^{a}}{r^{6}} \\ & +\sum_{b}\left[r^{3} \frac{\partial x^{a}}{\partial x^{b}}-3 x^{a} x^{b} r\right] \frac{J^{b}}{r^{6}}, \quad \text { (E.428) } \end{aligned}(×ζ)a=b[r3xbxb3(xb)2r]Jar6+b[r3xaxb3xaxbr]Jbr6, (E.428) 
from which the dipolar expression follows on doing the sum over b b bbb and translating back into vector notation.
(46.2) (c) For example, term 5 can be written as
h α β α β h μ ν = α ( h α β β h μ ν ) ( α h α β ) ( β h μ ν ) h α β α β h μ ν = α h α β β h μ ν α h α β β h μ ν h^(alpha beta)del_(alpha)del_(beta)h_(mu nu)=del_(alpha)(h^(alpha beta)del_(beta)h_(mu nu))-(del_(alpha)h^(alpha beta))(del_(beta)h_(mu nu))h^{\alpha \beta} \partial_{\alpha} \partial_{\beta} h_{\mu \nu}=\partial_{\alpha}\left(h^{\alpha \beta} \partial_{\beta} h_{\mu \nu}\right)-\left(\partial_{\alpha} h^{\alpha \beta}\right)\left(\partial_{\beta} h_{\mu \nu}\right)hαβαβhμν=α(hαββhμν)(αhαβ)(βhμν)
(E.429)
The first term on the right is a divergence which vanishes on the boundary. The second term is zero because α h α β = 0 α h α β = 0 del_(alpha)h^(alpha beta)=0\partial_{\alpha} h^{\alpha \beta}=0αhαβ=0. The other terms vanish through similar arguments.
(46.3) (c) The first term on the right-hand side of eqn 46.89 is a total derivative and we must have that T μ ν T μ ν T^(mu nu)T^{\mu \nu}Tμν and its time derivative vanish at the boundary of the mass.
(46.4) (b) Taking the derivative of part (a), we have
(E.430) h ˙ 1 R ( G m r ) 5 2 (E.430) h ˙ 1 R G m r 5 2 {:(E.430)h^(˙)~~(1)/(R)((Gm)/(r))^((5)/(2)):}\begin{equation*} \dot{h} \approx \frac{1}{R}\left(\frac{G m}{r}\right)^{\frac{5}{2}} \tag{E.430} \end{equation*}(E.430)h˙1R(Gmr)52
(d) For (i) take the solar-system mass to be the solar mass, and the size to be 60 astronomical units (A.U.) where one A.U. 1.5 × 10 11 m 1.5 × 10 11 m ~~1.5 xx10^(11)m\approx 1.5 \times 10^{11} \mathrm{~m}1.5×1011 m. (This is about twice the radius of Neptune's orbit.). Plugging in yields 5 kW 5 kW ~~5kW\approx 5 \mathrm{~kW}5 kW. For (ii), if we take the size of the system to be 4 r S 4 r S ~~4r_(S)\approx 4 r_{\mathrm{S}}4rS, then we obtain 10 49 W 10 49 W ~~10^(49)W\approx 10^{49} \mathrm{~W}1049 W. For (iii), taking M = 0.5 kg M = 0.5 kg M=0.5kgM=0.5 \mathrm{~kg}M=0.5 kg and R = 0.2 m R = 0.2 m R=0.2mR=0.2 \mathrm{~m}R=0.2 m, yields 10 81 W 10 81 W ~~10^(-81)W\approx 10^{-81} \mathrm{~W}1081 W.
(46.5) (c) Tr M k TT = M k k TT = ( P i k P j k 1 2 P k k P i j ) M i j = Tr M k TT = M k k TT = P i k P j k 1 2 P k k P i j M i j = TrM_(kℓ)^(TT)=M_(kk)^(TT)=(P_(ik)P_(jk)-(1)/(2)P_(kk)P_(ij))M_(ij)=\operatorname{Tr} M_{k \ell}^{\mathrm{TT}}=M_{k k}^{\mathrm{TT}}=\left(P_{i k} P_{j k}-\frac{1}{2} P_{k k} P_{i j}\right) M_{i j}=TrMkTT=MkkTT=(PikPjk12PkkPij)Mij= ( 1 1 2 P k k ) M i j = 0 1 1 2 P k k M i j = 0 (1-(1)/(2)P_(kk))M_(ij)=0\left(1-\frac{1}{2} P_{k k}\right) M_{i j}=0(112Pkk)Mij=0 since P k k = δ k k n k n k = 3 1 = 2 P k k = δ k k n k n k = 3 1 = 2 P_(kk)=delta_(kk)-n_(k)n_(k)=3-1=2P_{k k}=\delta_{k k}-n_{k} n_{k}=3-1=2Pkk=δkknknk=31=2. (e) Take n 1 = sin θ cos ϕ , n 2 = sin θ sin ϕ n 1 = sin θ cos ϕ , n 2 = sin θ sin ϕ n_(1)=sin theta cos phi,n_(2)=sin theta sin phin_{1}=\sin \theta \cos \phi, n_{2}=\sin \theta \sin \phin1=sinθcosϕ,n2=sinθsinϕ and n 3 = n 3 = n_(3)=n_{3}=n3= cos θ cos θ cos theta\cos \thetacosθ and d Ω = sin θ d θ d ϕ d Ω = sin θ d θ d ϕ dOmega=sin thetadthetadphi\mathrm{d} \Omega=\sin \theta \mathrm{d} \theta \mathrm{d} \phidΩ=sinθdθdϕ. Then you can show that n 1 n 2 d Ω = 0 , n 3 2 d Ω = 4 π / 3 , n 1 2 n 2 2 d Ω = 4 π / 15 n 1 n 2 d Ω = 0 , n 3 2 d Ω = 4 π / 3 , n 1 2 n 2 2 d Ω = 4 π / 15 intn_(1)n_(2)dOmega=0,intn_(3)^(2)dOmega=4pi//3,intn_(1)^(2)n_(2)^(2)dOmega=4pi//15\int n_{1} n_{2} \mathrm{~d} \Omega=0, \int n_{3}^{2} \mathrm{~d} \Omega=4 \pi / 3, \int n_{1}^{2} n_{2}^{2} \mathrm{~d} \Omega=4 \pi / 15n1n2 dΩ=0,n32 dΩ=4π/3,n12n22 dΩ=4π/15, n 1 n 2 n 3 2 d Ω = 0 n 1 n 2 n 3 2 d Ω = 0 intn_(1)n_(2)n_(3)^(2)dOmega=0\int n_{1} n_{2} n_{3}^{2} \mathrm{~d} \Omega=0n1n2n32 dΩ=0, and n 3 4 d Ω = 4 π / 5 n 3 4 d Ω = 4 π / 5 intn_(3)^(4)dOmega=4pi//5\int n_{3}^{4} \mathrm{~d} \Omega=4 \pi / 5n34 dΩ=4π/5. These are all the possibilities you need since other permutations of the possibilities you need since otry.
indices are related by symmetry
(46.6) (a) Using Newtonian mechanics, the force between the stars is G M 2 / 4 r 2 G M 2 / 4 r 2 GM^(2)//4r^(2)G M^{2} / 4 r^{2}GM2/4r2 and equating this to the centripetal force we obtain v 2 = G M / 4 r v 2 = G M / 4 r v^(2)=GM//4rv^{2}=G M / 4 rv2=GM/4r. Identifying ω = v / r ω = v / r omega=v//r\omega=v / rω=v/r, the answer follows.
(47.2) The combinations are
ϵ 1 i ϵ 2 = e i θ ( ϵ 1 i ϵ 2 ) , ϵ 3 = ϵ 3 , (E.431) ϵ 1 + i ϵ 2 = e i θ ( ϵ 1 + i ϵ 2 ) , ϵ 1 i ϵ 2 = e i θ ϵ 1 i ϵ 2 , ϵ 3 = ϵ 3 , (E.431) ϵ 1 + i ϵ 2 = e i θ ϵ 1 + i ϵ 2 , {:[epsilon_(1)-iepsilon_(2)=e^(itheta)(epsilon_(1)^(')-iepsilon_(2)^('))","],[epsilon_(3)=epsilon_(3)^(')","],[(E.431)epsilon_(1)+iepsilon_(2)=e^(-itheta)(epsilon_(1)^(')+iepsilon_(2)^('))","]:}\begin{align*} \epsilon_{1}-\mathrm{i} \epsilon_{2} & =\mathrm{e}^{\mathrm{i} \theta}\left(\epsilon_{1}^{\prime}-\mathrm{i} \epsilon_{2}^{\prime}\right), \\ \epsilon_{3} & =\epsilon_{3}^{\prime}, \\ \epsilon_{1}+\mathrm{i} \epsilon_{2} & =\mathrm{e}^{-\mathrm{i} \theta}\left(\epsilon_{1}^{\prime}+\mathrm{i} \epsilon_{2}^{\prime}\right), \tag{E.431} \end{align*}ϵ1iϵ2=eiθ(ϵ1iϵ2),ϵ3=ϵ3,(E.431)ϵ1+iϵ2=eiθ(ϵ1+iϵ2),
corresponding to h = 1 , 0 h = 1 , 0 h=1,0h=1,0h=1,0 and -1 , respectively.
(47.3) (b) The combinations are
ϵ 11 i ϵ 12 = e 2 i θ ( ϵ 11 i ϵ 12 ) (E.432) ϵ 11 + i ϵ 12 = e 2 i θ ( ϵ 11 + i ϵ 12 ) ϵ 11 i ϵ 12 = e 2 i θ ϵ 11 i ϵ 12 (E.432) ϵ 11 + i ϵ 12 = e 2 i θ ϵ 11 + i ϵ 12 {:[epsilon_(11)-iepsilon_(12)=e^(2itheta)(epsilon_(11)^(')-iepsilon_(12)^('))],[(E.432)epsilon_(11)+iepsilon_(12)=e^(-2itheta)(epsilon_(11)^(')+iepsilon_(12)^('))]:}\begin{align*} & \epsilon_{11}-\mathrm{i} \epsilon_{12}=\mathrm{e}^{2 \mathrm{i} \theta}\left(\epsilon_{11}^{\prime}-\mathrm{i} \epsilon_{12}^{\prime}\right) \\ & \epsilon_{11}+\mathrm{i} \epsilon_{12}=\mathrm{e}^{-2 \mathrm{i} \theta}\left(\epsilon_{11}^{\prime}+\mathrm{i} \epsilon_{12}^{\prime}\right) \tag{E.432} \end{align*}ϵ11iϵ12=e2iθ(ϵ11iϵ12)(E.432)ϵ11+iϵ12=e2iθ(ϵ11+iϵ12)
corresponding to h = 2 h = 2 h=2h=2h=2 and -2 , respectively.
where the separation is a = a 1 + a 2 = a 1 ( 1 + m 1 / m 2 ) a = a 1 + a 2 = a 1 1 + m 1 / m 2 a=a_(1)+a_(2)=a_(1)(1+m_(1)//m_(2))a=a_{1}+a_{2}=a_{1}\left(1+m_{1} / m_{2}\right)a=a1+a2=a1(1+m1/m2). The force on m 1 m 1 m_(1)m_{1}m1 obeys m 1 ω 2 a 1 = G m 1 m 2 / a 2 m 1 ω 2 a 1 = G m 1 m 2 / a 2 m_(1)omega^(2)a_(1)=Gm_(1)m_(2)//a^(2)m_{1} \omega^{2} a_{1}=G m_{1} m_{2} / a^{2}m1ω2a1=Gm1m2/a2, so ω 2 = G ( m 1 + m 2 ) / a 3 ω 2 = G m 1 + m 2 / a 3 omega^(2)=G(m_(1)+m_(2))//a^(3)\omega^{2}=G\left(m_{1}+m_{2}\right) / a^{3}ω2=G(m1+m2)/a3. The moment of inertia I I III is I = m 1 a 1 2 + m 2 a 2 = m 1 m 2 a 2 / ( m 1 + m 2 ) I = m 1 a 1 2 + m 2 a 2 = m 1 m 2 a 2 / m 1 + m 2 I=m_(1)a_(1)^(2)+m_(2)a^(2)=m_(1)m_(2)a^(2)//(m_(1)+m_(2))I=m_{1} a_{1}^{2}+m_{2} a^{2}=m_{1} m_{2} a^{2} /\left(m_{1}+m_{2}\right)I=m1a12+m2a2=m1m2a2/(m1+m2) after some algebra. The gravitational wave luminosity takes the form ( 32 G / 5 c 5 ) I 2 ω 6 32 G / 5 c 5 I 2 ω 6 (32 G//5c^(5))I^(2)omega^(6)\left(32 G / 5 c^{5}\right) I^{2} \omega^{6}(32G/5c5)I2ω6 and hence the required answer is obtained. Plugging in numbers gives about 200 W of power. The energy of one graviton is ω ω ℏomega\hbar \omegaω so this means about 10 43 10 43 10^(43)10^{43}1043 gravitons per second are emitted.
(b) Handwaving answer: Very roughly, using a mass 1 kg 1 kg ∼1kg\sim 1 \mathrm{~kg}1 kg, a length of 1 m 1 m ∼1m\sim 1 \mathrm{~m}1 m, and a frantic waving frequency of 2 Hz , so that ω 10 s 1 ω 10 s 1 omega∼10s^(-1)\omega \sim 10 \mathrm{~s}^{-1}ω10 s1, yields a luminosity of only of 2 Hz, so that ω 10 s 1 ω 10 s 1 omega∼10s^(-1)\omega \sim 10 \mathrm{~s}^{-1}ω10 s1, yields a luminosity of only
10 45 W 10 45 W ∼10^(-45)W\sim 10^{-45} \mathrm{~W}1045 W. The graviton energy is ω 10 33 J ω 10 33 J ℏomega∼10^(-33)J\hbar \omega \sim 10^{-33} \mathrm{~J}ω1033 J, so you will need to wave your hands for tens of thousands you will need to wave your hands for tens of
of years before you emit a single graviton.
(48.1) (a) The exterior derivative yields
(E.433) d ω 5 ^ = 0 + A α ^ x β ^ d x β ^ d x α ^ . (E.433) d ω 5 ^ = 0 + A α ^ x β ^ d x β ^ d x α ^ . {:(E.433)domega^( hat(5))=0+(delA_( hat(alpha)))/(delx^( hat(beta)))dx^( hat(beta))^^dx^( hat(alpha)).:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{5}}=0+\frac{\partial A_{\hat{\alpha}}}{\partial x^{\hat{\beta}}} \boldsymbol{d} x^{\hat{\beta}} \wedge \boldsymbol{d} x^{\hat{\alpha}} . \tag{E.433} \end{equation*}(E.433)dω5^=0+Aα^xβ^dxβ^dxα^.
Note that the derivative with respect to the fifth component vanishes, so we need only sum indices over the usual four dimensions. The expression can be rewritten as
(E.434) d ω 5 = 1 2 F α ^ β ^ d x α ^ d x β ^ (E.434) d ω 5 = 1 2 F α ^ β ^ d x α ^ d x β ^ {:(E.434)domega^(5)=(1)/(2)F_( hat(alpha) hat(beta))dx^( hat(alpha))^^dx^( hat(beta)):}\begin{equation*} \boldsymbol{d} \omega^{5}=\frac{1}{2} F_{\hat{\alpha} \hat{\beta}} \boldsymbol{d} x^{\hat{\alpha}} \wedge \boldsymbol{d} x^{\hat{\beta}} \tag{E.434} \end{equation*}(E.434)dω5=12Fα^β^dxα^dxβ^
from which the answer follows on identifying d x β ¯ = ω β ^ d x β ¯ = ω β ^ dx^( bar(beta))=omega^( hat(beta))\boldsymbol{d} x^{\bar{\beta}}=\boldsymbol{\omega}^{\hat{\beta}}dxβ¯=ωβ^. (b) Using idea 1, we have for the 1 -forms in fivedimensional space
d ω α ^ = ω a ^ α ^ ω a ^ = ω β ^ α ^ ω β ^ + ω 5 ^ α ^ ω 5 , (E.435) d ω α ^ = ω a ^ α ^ ω a ^ = ω β ^ α ^ ω β ^ + ω 5 ^ α ^ ω 5 ,  (E.435)  -domega^( hat(alpha))=omega_( hat(a))^( hat(alpha))^^omega^( hat(a))=omega_( hat(beta))^( hat(alpha))^^omega^( hat(beta))+omega_( hat(5))^( hat(alpha))^^omega^(5)," (E.435) "-\boldsymbol{d} \boldsymbol{\omega}^{\hat{\alpha}}=\boldsymbol{\omega}_{\hat{a}}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{\hat{a}}=\boldsymbol{\omega}_{\hat{\beta}}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{\hat{\beta}}+\boldsymbol{\omega}_{\hat{5}}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{5}, \text { (E.435) }dωα^=ωa^α^ωa^=ωβ^α^ωβ^+ω5^α^ω5, (E.435) 
where the sum over the Roman index a ^ a ^ hat(a)\hat{a}a^ is assumed to range over all five dimensions. From part (a) we have ω 5 ^ α ^ = 1 2 F α ^ β ^ ω β ^ ω 5 ^ α ^ = 1 2 F α ^ β ^ ω β ^ omega^( hat(5))_( hat(alpha))=(1)/(2)F_( hat(alpha) hat(beta))omega^( hat(beta))\boldsymbol{\omega}^{\hat{5}}{ }_{\hat{\alpha}}=\frac{1}{2} F_{\hat{\alpha} \hat{\beta}} \boldsymbol{\omega}^{\hat{\beta}}ω5^α^=12Fα^β^ωβ^, from which we can deduce ω 5 ^ α ^ = 1 2 F α ^ ω β ^ ω 5 ^ α ^ = 1 2 F α ^ ω β ^ omega_( hat(5))^( hat(alpha))=-(1)/(2)F^( hat(alpha))omega^( hat(beta))\boldsymbol{\omega}_{\hat{5}}^{\hat{\alpha}}=-\frac{1}{2} F^{\hat{\alpha}} \boldsymbol{\omega}^{\hat{\beta}}ω5^α^=12Fα^ωβ^. We therefore have
ω α ^ a ^ ω a ^ = ω α ^ β ^ ω β ^ + ω α ^ S ^ ω 5 ^ = ω α ^ β ^ ω β ^ 1 2 F α ^ β β ^ ω 5 ^ (E.436) = ( ω β ^ α ^ + 1 2 F β ^ α ^ ω 5 ^ ) ω β ^ . ω α ^ a ^ ω a ^ = ω α ^ β ^ ω β ^ + ω α ^ S ^ ω 5 ^ = ω α ^ β ^ ω β ^ 1 2 F α ^ β β ^ ω 5 ^ (E.436) = ω β ^ α ^ + 1 2 F β ^ α ^ ω 5 ^ ω β ^ . {:[omega^( hat(alpha))_( hat(a))^^omega^( hat(a))=omega^( hat(alpha))_( hat(beta))^^omega^( hat(beta))+omega^( hat(alpha))_( hat(S))^^omega^( hat(5))],[=omega^( hat(alpha))_( hat(beta))^^omega^( hat(beta))-(1)/(2)F^( hat(alpha))beta^( hat(beta))^^omega^( hat(5))],[(E.436)=(omega_( hat(beta))^( hat(alpha))+(1)/(2)F_( hat(beta))^( hat(alpha))omega^( hat(5)))^^omega^( hat(beta)).]:}\begin{align*} \boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{a}} \wedge \boldsymbol{\omega}^{\hat{a}} & =\boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{\beta}} \wedge \boldsymbol{\omega}^{\hat{\beta}}+\boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{S}} \wedge \boldsymbol{\omega}^{\hat{5}} \\ & =\boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{\beta}} \wedge \boldsymbol{\omega}^{\hat{\beta}}-\frac{1}{2} F^{\hat{\alpha}} \boldsymbol{\beta}^{\hat{\beta}} \wedge \boldsymbol{\omega}^{\hat{5}} \\ & =\left(\boldsymbol{\omega}_{\hat{\beta}}^{\hat{\alpha}}+\frac{1}{2} F_{\hat{\beta}}^{\hat{\alpha}} \boldsymbol{\omega}^{\hat{5}}\right) \wedge \boldsymbol{\omega}^{\hat{\beta}} . \tag{E.436} \end{align*}ωα^a^ωa^=ωα^β^ωβ^+ωα^S^ω5^=ωα^β^ωβ^12Fα^ββ^ω5^(E.436)=(ωβ^α^+12Fβ^α^ω5^)ωβ^.
In four-dimensional space, we have basis 1 -forms Ω α ^ Ω α ^ Omega^( hat(alpha))\boldsymbol{\Omega}^{\hat{\alpha}}Ωα^ and so we write d Ω α ^ = Ω α ^ β ^ Ω β ^ d Ω α ^ = Ω α ^ β ^ Ω β ^ -dOmega^( hat(alpha))=Omega^( hat(alpha))_( hat(beta))^^Omega^( hat(beta))-\boldsymbol{d} \boldsymbol{\Omega}^{\hat{\alpha}}=\boldsymbol{\Omega}^{\hat{\alpha}}{ }_{\hat{\beta}} \wedge \boldsymbol{\Omega}^{\hat{\beta}}dΩα^=Ωα^β^Ωβ^. However, it is also true that Ω α ^ = ω α ^ Ω α ^ = ω α ^ Omega^( hat(alpha))=omega^( hat(alpha))\boldsymbol{\Omega}^{\hat{\alpha}}=\boldsymbol{\omega}^{\hat{\alpha}}Ωα^=ωα^, so d ω α ^ = d Ω α ^ d ω α ^ = d Ω α ^ domega^( hat(alpha))=dOmega^( hat(alpha))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\alpha}}=\boldsymbol{d} \boldsymbol{\Omega}^{\hat{\alpha}}dωα^=dΩα^, and we conclude
(E.437) Ω β ^ α ^ = ω β ^ α ^ + 1 2 F β ^ α ^ ω 5 ^ , (E.437) Ω β ^ α ^ = ω β ^ α ^ + 1 2 F β ^ α ^ ω 5 ^ , {:(E.437)Omega_( hat(beta))^( hat(alpha))=omega_( hat(beta))^( hat(alpha))+(1)/(2)F_( hat(beta))^( hat(alpha))omega^( hat(5))",":}\begin{equation*} \boldsymbol{\Omega}_{\hat{\beta}}^{\hat{\alpha}}=\boldsymbol{\omega}_{\hat{\beta}}^{\hat{\alpha}}+\frac{1}{2} F_{\hat{\beta}}^{\hat{\alpha}} \boldsymbol{\omega}^{\hat{5}}, \tag{E.437} \end{equation*}(E.437)Ωβ^α^=ωβ^α^+12Fβ^α^ω5^,
which can be rearranged to give the answer.
(c) The ingredients are
d ω β ^ α ^ = d Ω α ^ β ^ 1 2 F α ^ β ^ , γ ^ ( ω γ ^ ω 5 ^ ) 1 2 F α ^ β ^ d ω 5 ^ , ω α ^ μ ^ ω β ^ μ ^ = ( Ω μ ^ α ^ 1 2 F μ ^ α ^ ω 5 ) ( Ω β ^ μ ^ 1 2 F β ^ β ^ μ ^ ω 5 ) , (E.439) and (E.440) ω α ^ 5 ^ ω 5 ^ β ^ = ( 1 2 F γ ^ α ^ ω γ ^ ) ( 1 2 F β ^ δ ^ ω δ ^ ) . (E.440) d ω β ^ α ^ = d Ω α ^ β ^ 1 2 F α ^ β ^ , γ ^ ω γ ^ ω 5 ^ 1 2 F α ^ β ^ d ω 5 ^ , ω α ^ μ ^ ω β ^ μ ^ = Ω μ ^ α ^ 1 2 F μ ^ α ^ ω 5 Ω β ^ μ ^ 1 2 F β ^ β ^ μ ^ ω 5 , (E.439)   and  (E.440) ω α ^ 5 ^ ω 5 ^ β ^ = 1 2 F γ ^ α ^ ω γ ^ 1 2 F β ^ δ ^ ω δ ^ .  (E.440)  {:[qquad domega_( hat(beta))^( hat(alpha))=dOmega^( hat(alpha))_( hat(beta))-(1)/(2)F^( hat(alpha))_( hat(beta), hat(gamma))(omega^( hat(gamma))^^omega^( hat(5)))-(1)/(2)F^( hat(alpha))_( hat(beta))domega^( hat(5))","],[omega^( hat(alpha))_( hat(mu))^^omega_( hat(beta))^( hat(mu))=(Omega_( hat(mu))^( hat(alpha))-(1)/(2)F_( hat(mu))^( hat(alpha))omega^(5))^^(Omega_( hat(beta))^( hat(mu))-(1)/(2)F_({:( hat(beta))_( hat(beta))^( hat(mu))omega^(5)),)^((E.439) ):}],[" and "],[(E.440)omega^( hat(alpha))_( hat(5))^^omega^( hat(5))_( hat(beta))=(-(1)/(2)F_( hat(gamma))^( hat(alpha))omega^( hat(gamma)))^^((1)/(2)F_( hat(beta) hat(delta))omega^( hat(delta))).quad" (E.440) "]:}\begin{align*} & \qquad \boldsymbol{d} \boldsymbol{\omega}_{\hat{\beta}}^{\hat{\alpha}}=\boldsymbol{d} \boldsymbol{\Omega}^{\hat{\alpha}}{ }_{\hat{\beta}}-\frac{1}{2} F^{\hat{\alpha}}{ }_{\hat{\beta}, \hat{\gamma}}\left(\boldsymbol{\omega}^{\hat{\gamma}} \wedge \boldsymbol{\omega}^{\hat{5}}\right)-\frac{1}{2} F^{\hat{\alpha}}{ }_{\hat{\beta}} \boldsymbol{d} \omega^{\hat{5}}, \\ & \boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{\mu}} \wedge \boldsymbol{\omega}_{\hat{\beta}}^{\hat{\mu}}=\left(\boldsymbol{\Omega}_{\hat{\mu}}^{\hat{\alpha}}-\frac{1}{2} F_{\hat{\mu}}^{\hat{\alpha}} \boldsymbol{\omega}^{5}\right) \wedge\left(\boldsymbol{\Omega}_{\hat{\beta}}^{\hat{\mu}}-\frac{1}{2} F_{\left.\hat{\beta}{ }_{\hat{\beta}}^{\hat{\mu}} \boldsymbol{\omega}^{5}\right),}^{\text {(E.439) }}\right. \\ & \text { and } \\ & \boldsymbol{\omega}^{\hat{\alpha}}{ }_{\hat{5}} \wedge \boldsymbol{\omega}^{\hat{5}}{ }_{\hat{\beta}}=\left(-\frac{1}{2} F_{\hat{\gamma}}^{\hat{\alpha}} \boldsymbol{\omega}^{\hat{\gamma}}\right) \wedge\left(\frac{1}{2} F_{\hat{\beta} \hat{\delta}} \boldsymbol{\omega}^{\hat{\delta}}\right) . \quad \text { (E.440) } \tag{E.440} \end{align*}dωβ^α^=dΩα^β^12Fα^β^,γ^(ωγ^ω5^)12Fα^β^dω5^,ωα^μ^ωβ^μ^=(Ωμ^α^12Fμ^α^ω5)(Ωβ^μ^12Fβ^β^μ^ω5),(E.439)  and (E.440)ωα^5^ω5^β^=(12Fγ^α^ωγ^)(12Fβ^δ^ωδ^). (E.440) 
(f) The ingredients are
(E.441) d ω 5 ^ α ^ = d ( 1 2 F α ^ β ^ ω β ^ ) , (E.442) and ω 5 ^ β ^ ω β ^ α ^ = 1 2 F β ^ γ ^ ω γ ^ ( Ω β ^ α ^ 1 2 F α ^ β ^ ω 5 ^ ) (E.442) (E.441) d ω 5 ^ α ^ = d 1 2 F α ^ β ^ ω β ^ , (E.442)  and  ω 5 ^ β ^ ω β ^ α ^ = 1 2 F β ^ γ ^ ω γ ^ Ω β ^ α ^ 1 2 F α ^ β ^ ω 5 ^  (E.442)  {:[(E.441)domega^( hat(5))_( hat(alpha))=d((1)/(2)F_( hat(alpha) hat(beta))omega^( hat(beta)))","],[(E.442)" and "],[omega^( hat(5))_( hat(beta))^^omega^( hat(beta))_( hat(alpha))=(1)/(2)F_( hat(beta) hat(gamma))omega^( hat(gamma))^^(Omega^( hat(beta))_( hat(alpha))-(1)/(2)F_( hat(alpha))^( hat(beta))omega^( hat(5)))*" (E.442) "]:}\begin{gather*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{5}}{ }_{\hat{\alpha}}=\boldsymbol{d}\left(\frac{1}{2} F_{\hat{\alpha} \hat{\beta}} \boldsymbol{\omega}^{\hat{\beta}}\right), \tag{E.441}\\ \text { and } \tag{E.442}\\ \boldsymbol{\omega}^{\hat{5}}{ }_{\hat{\beta}} \wedge \boldsymbol{\omega}^{\hat{\beta}}{ }_{\hat{\alpha}}=\frac{1}{2} F_{\hat{\beta} \hat{\gamma}} \boldsymbol{\omega}^{\hat{\gamma}} \wedge\left(\boldsymbol{\Omega}^{\hat{\beta}}{ }_{\hat{\alpha}}-\frac{1}{2} F_{\hat{\alpha}}^{\hat{\beta}} \boldsymbol{\omega}^{\hat{5}}\right) \cdot \text { (E.442) } \end{gather*}(E.441)dω5^α^=d(12Fα^β^ωβ^),(E.442) and ω5^β^ωβ^α^=12Fβ^γ^ωγ^(Ωβ^α^12Fα^β^ω5^) (E.442) 
(49.1) The wavelength λ λ lambda\lambdaλ can be estimated using E = h c / λ E = h c / λ E=hc//lambdaE=h c / \lambdaE=hc/λ and this yields λ P λ P lambda∼ℓ_(P)\lambda \sim \ell_{\mathrm{P}}λP, ignoring small numerical constants. The gravitational self-energy for a particle of mass m P m P m_(P)m_{P}mP and 'size' λ λ ∼lambda\sim \lambdaλ is G m P 2 / λ G m P 2 / λ ~~Gm_(P)^(2)//lambda\approx G m_{\mathrm{P}}^{2} / \lambdaGmP2/λ, which also comes out to be E P E P ∼E_(P)\sim E_{\mathrm{P}}EP. By the same argument, this particle would have a Compton wavelength h / m P c h / m P c h//m_(P)ch / m_{\mathrm{P}} ch/mPc and Schwarzschild radius 2 G m P / c 2 2 G m P / c 2 2Gm_(P)//c^(2)2 G m_{\mathrm{P}} / \mathrm{c}^{2}2GmP/c2 would also be of the order of P P ℓ_(P)\ell_{\mathrm{P}}P.
(49.2) From eqn 49.17 the action is proportional to the area. Recall that we showed that elements formed by combinations like d τ d σ γ d τ d σ γ dtaudsigmasqrt(-gamma)\mathrm{d} \tau \mathrm{d} \sigma \sqrt{-\gamma}dτdσγ are invariants, meaning that we'll obtain the same answer if we use different coordinates. This is just what we mean by reparametrizing the string.
(49.5) (a) The Lagrangian is
Lagrangian is (E.443) L = T 0 [ ( X σ X t ) 2 + ( X σ ) 2 ( X σ ) 2 ( X t ) 2 ] 1 2  Lagrangian is  (E.443) L = T 0 X σ X t 2 + X σ 2 X σ 2 X t 2 1 2 {:[" Lagrangian is "],[(E.443){:[L=-T_(0)[((del( vec(X)))/(del sigma)*(del( vec(X)))/(del t))^(2)+((del( vec(X)))/(del sigma))^(2):}],[-((del( vec(X)))/(del sigma))^(2)((del( vec(X)))/(del t))^(2)]^((1)/(2))]:}]:}\begin{align*} \text { Lagrangian is } \\ \begin{aligned} \mathcal{L}= & -T_{0}\left[\left(\frac{\partial \vec{X}}{\partial \sigma} \cdot \frac{\partial \vec{X}}{\partial t}\right)^{2}+\left(\frac{\partial \vec{X}}{\partial \sigma}\right)^{2}\right. \\ & \left.-\left(\frac{\partial \vec{X}}{\partial \sigma}\right)^{2}\left(\frac{\partial \vec{X}}{\partial t}\right)^{2}\right]^{\frac{1}{2}} \end{aligned} \tag{E.443} \end{align*} Lagrangian is (E.443)L=T0[(XσXt)2+(Xσ)2(Xσ)2(Xt)2]12
The components of the momentum are
P i = L ( t X i ) P i = L t X i P^(i)=(delL)/(del(del_(t)X^(i)))P^{i}=\frac{\partial \mathcal{L}}{\partial\left(\partial_{t} X^{i}\right)}Pi=L(tXi)
(E.444) = T 0 2 L [ X i t X σ X i σ ( X σ X t ) ] (E.444) = T 0 2 L X i t X σ X i σ X σ X t {:(E.444)=(T_(0)^(2))/(L)[(delX^(i))/(del t)*(del( vec(X)))/(del sigma)-(delX^(i))/(del sigma)*((del( vec(X)))/(del sigma)*(del( vec(X)))/(del t))]:}\begin{equation*} =\frac{T_{0}^{2}}{\mathcal{L}}\left[\frac{\partial X^{i}}{\partial t} \cdot \frac{\partial \vec{X}}{\partial \sigma}-\frac{\partial X^{i}}{\partial \sigma} \cdot\left(\frac{\partial \vec{X}}{\partial \sigma} \cdot \frac{\partial \vec{X}}{\partial t}\right)\right] \tag{E.444} \end{equation*}(E.444)=T02L[XitXσXiσ(XσXt)]
If we then use X i / σ = ( d s / d σ ) ( X i / s ) X i / σ = ( d s / d σ ) X i / s delX^(i)//del sigma=(ds//dsigma)(delX^(i)//del s)\partial X^{i} / \partial \sigma=(\mathrm{d} s / \mathrm{d} \sigma)\left(\partial X^{i} / \partial s\right)Xi/σ=(ds/dσ)(Xi/s) and ( X / s ) 2 = 1 ( X / s ) 2 = 1 (del vec(X)//del s)^(2)=1(\partial \vec{X} / \partial s)^{2}=1(X/s)2=1, we have
(E.445) L = T 0 ( 1 v 2 ) 1 2 d s d σ , (E.445) L = T 0 1 v 2 1 2 d s d σ , {:(E.445)L=-T_(0)(1-v_(_|_)^(2))^((1)/(2))((d)s)/((d)sigma)",":}\begin{equation*} \mathcal{L}=-T_{0}\left(1-v_{\perp}^{2}\right)^{\frac{1}{2}} \frac{\mathrm{~d} s}{\mathrm{~d} \sigma}, \tag{E.445} \end{equation*}(E.445)L=T0(1v2)12 ds dσ,
and the momentum can then be expressed as
(E.446) P = T 0 v 1 v 2 d s d σ (E.446) P = T 0 v 1 v 2 d s d σ {:(E.446) vec(P)=T_(0)( vec(v)_(_|_))/(sqrt(1-v_(_|_)^(2)))((d)s)/((d)sigma):}\begin{equation*} \vec{P}=T_{0} \frac{\vec{v}_{\perp}}{\sqrt{1-v_{\perp}^{2}}} \frac{\mathrm{~d} s}{\mathrm{~d} \sigma} \tag{E.446} \end{equation*}(E.446)P=T0v1v2 ds dσ
(b) The Hamiltonian density is given by
(E.447) H = P X t L (E.447) H = P X t L {:(E.447)H= vec(P)*(del( vec(X)))/(del t)-L:}\begin{equation*} \mathcal{H}=\vec{P} \cdot \frac{\partial \vec{X}}{\partial t}-\mathcal{L} \tag{E.447} \end{equation*}(E.447)H=PXtL
Using the property that
(E.448) v X t = v 2 (E.448) v X t = v 2 {:(E.448) vec(v)_(_|_)*(del( vec(X)))/(del t)=v_(_|_)^(2):}\begin{equation*} \vec{v}_{\perp} \cdot \frac{\partial \vec{X}}{\partial t}=v_{\perp}^{2} \tag{E.448} \end{equation*}(E.448)vXt=v2
we find
(E.449) H = T 0 d s d σ 1 1 v 2 (E.449) H = T 0 d s d σ 1 1 v 2 {:(E.449)H=T_(0)((d)s)/((d)sigma)(1)/(sqrt(1-v_(_|_)^(2))):}\begin{equation*} \mathcal{H}=T_{0} \frac{\mathrm{~d} s}{\mathrm{~d} \sigma} \frac{1}{\sqrt{1-v_{\perp}^{2}}} \tag{E.449} \end{equation*}(E.449)H=T0 ds dσ11v2
The Hamiltonian can then be written as
(E.450) H = d σ H = T 0 d s 1 v 2 (E.450) H = d σ H = T 0 d s 1 v 2 {:(E.450)H=intdsigmaH=int(T_(0)(d)s)/(sqrt(1-v_(_|_)^(2))):}\begin{equation*} H=\int \mathrm{d} \sigma \mathcal{H}=\int \frac{T_{0} \mathrm{~d} s}{\sqrt{1-v_{\perp}^{2}}} \tag{E.450} \end{equation*}(E.450)H=dσH=T0 ds1v2
This has the form of E = m / 1 v 2 E = m / 1 v 2 E=m//sqrt(1-v^(2))E=m / \sqrt{1-v^{2}}E=m/1v2 for transverse motion, with a rest mass given by the string tension, so makes sense if interpreted as an energy.
(49.7) (a) The computation follows, e.g. Example 36.4. We obtain R θ ^ θ ^ = R ϕ ^ ϕ ^ = 1 / r 0 2 R θ ^ θ ^ = R ϕ ^ ϕ ^ = 1 / r 0 2 R_( hat(theta) hat(theta))=R_( hat(phi) hat(phi))=1//r_(0)^(2)R_{\hat{\theta} \hat{\theta}}=R_{\hat{\phi} \hat{\phi}}=1 / r_{0}^{2}Rθ^θ^=Rϕ^ϕ^=1/r02 and R = 2 / r 0 2 R = 2 / r 0 2 R=2//r_(0)^(2)R=2 / r_{0}^{2}R=2/r02. The Einstein equation is solved with 8 π ρ = 1 / r 0 2 8 π ρ = 1 / r 0 2 8pi rho=1//r_(0)^(2)8 \pi \rho=1 / r_{0}^{2}8πρ=1/r02.
(b) The cross-sectional area of the string is
0 θ m r 0 d θ 0 2 π r 0 sin θ d ϕ = 2 π r 0 2 ( 1 cos θ m ) 0 θ m r 0 d θ 0 2 π r 0 sin θ d ϕ = 2 π r 0 2 1 cos θ m int_(0)^(theta_(m))r_(0)dthetaint_(0)^(2pi)r_(0)sin thetadphi=2pir_(0)^(2)(1-cos theta_(m))\int_{0}^{\theta_{\mathrm{m}}} r_{0} \mathrm{~d} \theta \int_{0}^{2 \pi} r_{0} \sin \theta \mathrm{~d} \phi=2 \pi r_{0}^{2}\left(1-\cos \theta_{\mathrm{m}}\right)0θmr0 dθ02πr0sinθ dϕ=2πr02(1cosθm). (E.451)
The mass per unit length is then 2 π r 0 2 ( 1 cos θ m ) ρ 2 π r 0 2 1 cos θ m ρ 2pir_(0)^(2)(1-cos theta_(m))rho2 \pi r_{0}^{2}\left(1-\cos \theta_{\mathrm{m}}\right) \rho2πr02(1cosθm)ρ or, using the previous part, ( 1 cos θ m ) / 4 1 cos θ m / 4 (1-cos theta_(m))//4\left(1-\cos \theta_{\mathrm{m}}\right) / 4(1cosθm)/4.
(49.8) (a) The flat cylindrical metric follows from the substitution
(E.452) r = r 0 sin θ cos θ m , ϕ = ϕ cos θ m (E.452) r = r 0 sin θ cos θ m , ϕ = ϕ cos θ m {:(E.452)r^(')=r_(0)(sin theta)/(cos theta_(m))","quadphi^(')=phi cos theta_(m):}\begin{equation*} r^{\prime}=r_{0} \frac{\sin \theta}{\cos \theta_{\mathrm{m}}}, \quad \phi^{\prime}=\phi \cos \theta_{\mathrm{m}} \tag{E.452} \end{equation*}(E.452)r=r0sinθcosθm,ϕ=ϕcosθm
(b) The new variable ϕ ϕ phi^(')\phi^{\prime}ϕ has a range 0 ϕ 2 π cos θ m 0 ϕ 2 π cos θ m 0 <= phi^(') <= 2pi cos theta_(m)0 \leq \phi^{\prime} \leq 2 \pi \cos \theta_{\mathrm{m}}0ϕ2πcosθm. As a result, we have d ϕ g ϕ ϕ = 2 π a cos θ m d ϕ g ϕ ϕ = 2 π a cos θ m intdphi^(')sqrt(g_(phi^(')phi^(')))=2pi a cos theta_(m)\int \mathrm{d} \phi^{\prime} \sqrt{g_{\phi^{\prime} \phi^{\prime}}}=2 \pi a \cos \theta_{\mathrm{m}}dϕgϕϕ=2πacosθm. Despite the spacetime looking flat, the fact that this circumference is less than 2 π a 2 π a 2pi a2 \pi a2πa gives a sense of the curvature caused by the string.
(49.9) L 1 = 1 2 a × b , L 2 = 1 2 b × c , L 3 = 1 2 c × a L 1 = 1 2 a × b , L 2 = 1 2 b × c , L 3 = 1 2 c × a vec(L)_(1)=(1)/(2) vec(a)xx vec(b), vec(L)_(2)=(1)/(2) vec(b)xx vec(c), vec(L)_(3)=(1)/(2) vec(c)xx vec(a)\vec{L}_{1}=\frac{1}{2} \vec{a} \times \vec{b}, \vec{L}_{2}=\frac{1}{2} \vec{b} \times \vec{c}, \vec{L}_{3}=\frac{1}{2} \vec{c} \times \vec{a}L1=12a×b,L2=12b×c,L3=12c×a, and L 4 = 1 2 ( b c ) × ( a c ) = 1 2 ( b × a c × a b × c ) L 4 = 1 2 ( b c ) × ( a c ) = 1 2 ( b × a c × a b × c ) vec(L)_(4)=(1)/(2)( vec(b)- vec(c))xx( vec(a)- vec(c))=(1)/(2)( vec(b)xx vec(a)- vec(c)xx vec(a)- vec(b)xx vec(c))\vec{L}_{4}=\frac{1}{2}(\vec{b}-\vec{c}) \times(\vec{a}-\vec{c})=\frac{1}{2}(\vec{b} \times \vec{a}-\vec{c} \times \vec{a}-\vec{b} \times \vec{c})L4=12(bc)×(ac)=12(b×ac×ab×c), and so the closure property is trivially satisfied.
(49.10) A tetrahedron is one sixth of the volume of the parallelepiped generated by a × b c a × b c vec(a)xx vec(b)* vec(c)\vec{a} \times \vec{b} \cdot \vec{c}a×bc. The modulus sign is because the scalar triple product can be negative debecause the scalar triple product can be negative de-
pending on how the vectors are oriented, but volume pending on how the vectors are oriented, but volume
can only be positive. Substitution of the results from can only be positive. Sub
the previous problem give
L 1 × L 2 L 3 = 1 8 [ ( a × b ) × ( b × c ) ( c × a ) ] , ( E .453 ) L 1 × L 2 L 3 = 1 8 [ ( a × b ) × ( b × c ) ( c × a ) ] , ( E .453 ) vec(L)_(1)xx vec(L)_(2)* vec(L)_(3)=(1)/(8)[( vec(a)xx vec(b))xx( vec(b)xx vec(c))*( vec(c)xx vec(a))],quad(E.453)\vec{L}_{1} \times \vec{L}_{2} \cdot \vec{L}_{3}=\frac{1}{8}[(\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})], \quad(\mathrm{E} .453)L1×L2L3=18[(a×b)×(b×c)(c×a)],(E.453)
and using the vector identity
(E.454) A × ( B × C ) = ( A C ) B ( A B ) C (E.454) A × ( B × C ) = ( A C ) B ( A B ) C {:(E.454) vec(A)xx( vec(B)xx vec(C))=( vec(A)* vec(C)) vec(B)-( vec(A)* vec(B)) vec(C):}\begin{equation*} \vec{A} \times(\vec{B} \times \vec{C})=(\vec{A} \cdot \vec{C}) \vec{B}-(\vec{A} \cdot \vec{B}) \vec{C} \tag{E.454} \end{equation*}(E.454)A×(B×C)=(AC)B(AB)C
the result follows.
(49.12) Substitute the given coordinates into the line element d s 2 = d T 2 + d X 2 + d Y 2 d W 2 d s 2 = d T 2 + d X 2 + d Y 2 d W 2 ds^(2)=-dT^(2)+dX^(2)+dY^(2)-dW^(2)\mathrm{d} s^{2}=-\mathrm{d} T^{2}+\mathrm{d} X^{2}+\mathrm{d} Y^{2}-\mathrm{d} W^{2}ds2=dT2+dX2+dY2dW2. The answer follows after several lines of algebra.
(50.1) (a) P μ ν = g μ ν + u μ u ν P μ ν = g μ ν + u μ u ν P_(mu nu)=g_(mu nu)+u_(mu)u_(nu)P_{\mu \nu}=g_{\mu \nu}+u_{\mu} u_{\nu}Pμν=gμν+uμuν.
(b) In components, we have P μ ν v ν = v μ + u μ u ν v ν P μ ν v ν = v μ + u μ u ν v ν P_(mu nu)v^(nu)=v_(mu)+u_(mu)u_(nu)v^(nu)P_{\mu \nu} v^{\nu}=v_{\mu}+u_{\mu} u_{\nu} v^{\nu}Pμνvν=vμ+uμuνvν. Then
P μ ν v ν u μ = v μ u μ + u μ u ν v ν u μ (E.455) = v μ u μ u ν v ν = 0 P μ ν v ν u μ = v μ u μ + u μ u ν v ν u μ (E.455) = v μ u μ u ν v ν = 0 {:[P_(mu nu)v^(nu)u^(mu)=v_(mu)u^(mu)+u_(mu)u_(nu)v^(nu)u^(mu)],[(E.455)=v_(mu)u^(mu)-u_(nu)v^(nu)=0]:}\begin{align*} P_{\mu \nu} v^{\nu} u^{\mu} & =v_{\mu} u^{\mu}+u_{\mu} u_{\nu} v^{\nu} u^{\mu} \\ & =v_{\mu} u^{\mu}-u_{\nu} v^{\nu}=0 \tag{E.455} \end{align*}Pμνvνuμ=vμuμ+uμuνvνuμ(E.455)=vμuμuνvν=0
since u μ u μ = 1 u μ u μ = 1 u_(mu)u^(mu)=-1u_{\mu} u^{\mu}=-1uμuμ=1.
(c) P μ ν P μ ν = g μ ν ( g μ ν u μ u ν ) + u μ u ν ( g μ ν u μ u ν ) = 4 P μ ν P μ ν = g μ ν g μ ν u μ u ν + u μ u ν g μ ν u μ u ν = 4 P^(mu nu)P_(mu nu)=g^(mu nu)(g_(mu nu)u_(mu)u_(nu))+u^(mu)u^(nu)(g_(mu nu)u_(mu)u_(nu))=4-P^{\mu \nu} P_{\mu \nu}=g^{\mu \nu}\left(g_{\mu \nu} u_{\mu} u_{\nu}\right)+u^{\mu} u^{\nu}\left(g_{\mu \nu} u_{\mu} u_{\nu}\right)=4-PμνPμν=gμν(gμνuμuν)+uμuν(gμνuμuν)=4 1 1 + 1 = 3 1 1 + 1 = 3 1-1+1=31-1+1=311+1=3.
(d) P α β u α ; β = g α β u α ; β + u α u β u α ; β P α β u α ; β = g α β u α ; β + u α u β u α ; β P^(alpha beta)u_(alpha;beta)=g^(alpha beta)u_(alpha;beta)+u^(alpha)u^(beta)u_(alpha;beta)P^{\alpha \beta} u_{\alpha ; \beta}=g^{\alpha \beta} u_{\alpha ; \beta}+u^{\alpha} u^{\beta} u_{\alpha ; \beta}Pαβuα;β=gαβuα;β+uαuβuα;β for a geodesic. The second term is ( u u ) u = 0 u u u = 0 (grad_(u)u)*u=0\left(\nabla_{\boldsymbol{u}} \boldsymbol{u}\right) \cdot \boldsymbol{u}=0(uu)u=0. The first term is u β ; β = u u β ; β = u u^(beta)_(;beta)=grad*uu^{\beta}{ }_{; \beta}=\boldsymbol{\nabla} \cdot \boldsymbol{u}uβ;β=u.
(e) We find
(E.456) P ( n , v ) = n v | n | 2 n v = 0 (E.456) P ( n , v ) = n v | n | 2 n v = 0 {:(E.456)P(n","v)=n*v-|n|^(2)n*v=0:}\begin{equation*} P(\boldsymbol{n}, \boldsymbol{v})=\boldsymbol{n} \cdot \boldsymbol{v}-|\boldsymbol{n}|^{2} \boldsymbol{n} \cdot \boldsymbol{v}=0 \tag{E.456} \end{equation*}(E.456)P(n,v)=nv|n|2nv=0
if n n = 1 n n = 1 n*n=1\boldsymbol{n} \cdot \boldsymbol{n}=1nn=1.
(50.2) (a) The vector W W W\boldsymbol{W}W which links geodesics is Lie dragged, so we have £ u W = [ u , W ] = 0 £ u W = [ u , W ] = 0 £_(u)W=[u,W]=0£_{\boldsymbol{u}} \boldsymbol{W}=[\boldsymbol{u}, \boldsymbol{W}]=0£uW=[u,W]=0, and so
(E.457) u W = W u (E.457) u W = W u {:(E.457)grad_(u)W=grad_(W)u:}\begin{equation*} \nabla_{u} \boldsymbol{W}=\boldsymbol{\nabla}_{W} \boldsymbol{u} \tag{E.457} \end{equation*}(E.457)uW=Wu
We can therefore write
( u W ) μ = ( W u ) μ = W ν ( ν u ) μ (E.458) = u μ ; ν W ν u W μ = W u μ = W ν ν u μ (E.458) = u μ ; ν W ν {:[(grad_(u)W)^(mu)=(grad_(W)u)^(mu)=W^(nu)(grad_(nu)u)^(mu)],[(E.458)=u^(mu)_(;nu)W^(nu)]:}\begin{align*} \left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{W}\right)^{\mu} & =\left(\boldsymbol{\nabla}_{\boldsymbol{W}} \boldsymbol{u}\right)^{\mu}=W^{\nu}\left(\boldsymbol{\nabla}_{\nu} \boldsymbol{u}\right)^{\mu} \\ & =u^{\mu}{ }_{; \nu} W^{\nu} \tag{E.458} \end{align*}(uW)μ=(Wu)μ=Wν(νu)μ(E.458)=uμ;νWν
(b) We write u μ u μ ; ν = 1 2 ( u μ u μ ) ; ν = 0 u μ u μ ; ν = 1 2 u μ u μ ; ν = 0 u^(mu)u_(mu;nu)=(1)/(2)(u^(mu)u_(mu))_(;nu)=0u^{\mu} u_{\mu ; \nu}=\frac{1}{2}\left(u^{\mu} u_{\mu}\right)_{; \nu}=0uμuμ;ν=12(uμuμ);ν=0, since u μ u μ = u μ u μ = u^(mu)u_(mu)=u^{\mu} u_{\mu}=uμuμ= -1. The other expression u μ ; ν u ν = 0 u μ ; ν u ν = 0 u_(mu;nu)u^(nu)=0u_{\mu ; \nu} u^{\nu}=0uμ;νuν=0 follows from the fact that u u u\boldsymbol{u}u is tangent to a geodesic.
(c) We have
(E.459) ( u B ) μ ν = u α u μ ; ν α . (E.459) u B μ ν = u α u μ ; ν α . {:(E.459)(grad_(u)B)_(mu nu)=u^(alpha)u_(mu;nu alpha).:}\begin{equation*} \left(\nabla_{u} \boldsymbol{B}\right)_{\mu \nu}=u^{\alpha} u_{\mu ; \nu \alpha} . \tag{E.459} \end{equation*}(E.459)(uB)μν=uαuμ;να.
Note first that
(E.460) u μ ; ν α = u μ ; α ν + R μ ν α β u β , (E.460) u μ ; ν α = u μ ; α ν + R μ ν α β u β , {:(E.460)u_(mu;nu alpha)=u_(mu;alpha nu)+R_(mu nu alpha)^(beta)u_(beta)",":}\begin{equation*} u_{\mu ; \nu \alpha}=u_{\mu ; \alpha \nu}+R_{\mu \nu \alpha}^{\beta} u_{\beta}, \tag{E.460} \end{equation*}(E.460)uμ;να=uμ;αν+Rμναβuβ,
so we can write
(E.461) ( u B ) μ ν = u α ( u μ ; α ν + R μ ν α β u β ) . (E.461) u B μ ν = u α u μ ; α ν + R μ ν α β u β . {:(E.461)(grad_(u)B)_(mu nu)=u^(alpha)(u_(mu;alpha nu)+R_(mu nu alpha)^(beta)u_(beta)).:}\begin{equation*} \left(\nabla_{u} \boldsymbol{B}\right)_{\mu \nu}=u^{\alpha}\left(u_{\mu ; \alpha \nu}+R_{\mu \nu \alpha}^{\beta} u_{\beta}\right) . \tag{E.461} \end{equation*}(E.461)(uB)μν=uα(uμ;αν+Rμναβuβ).
Consider the first term on the right, which we can cast as
(E.462) u α u μ ; α ν = ( u α u μ ; α ) ; ν u α ; ν u μ ; α = u α ; ν u μ ; α . (E.462) u α u μ ; α ν = u α u μ ; α ; ν u α ; ν u μ ; α = u α ; ν u μ ; α . {:(E.462)u^(alpha)u_(mu;alpha nu)=(u^(alpha)u_(mu;alpha))_(;nu)-u^(alpha)_(;nu)u_(mu;alpha)=u^(alpha)_(;nu)u_(mu;alpha).:}\begin{equation*} u^{\alpha} u_{\mu ; \alpha \nu}=\left(u^{\alpha} u_{\mu ; \alpha}\right)_{; \nu}-u^{\alpha}{ }_{; \nu} u_{\mu ; \alpha}=u^{\alpha}{ }_{; \nu} u_{\mu ; \alpha} . \tag{E.462} \end{equation*}(E.462)uαuμ;αν=(uαuμ;α);νuα;νuμ;α=uα;νuμ;α.
The second term can be rearranged
R μ ν α β u β u α = R μ α ν β u β u α (E.463) = R β μ α ν u β u α . R μ ν α β u β u α = R μ α ν β u β u α (E.463) = R β μ α ν u β u α . {:[R_(mu nu alpha)^(beta)u_(beta)u^(alpha)=-R_(mu alpha nu)^(beta)u_(beta)u^(alpha)],[(E.463)=-R_(beta mu alpha nu)u^(beta)u^(alpha).]:}\begin{align*} R_{\mu \nu \alpha}^{\beta} u_{\beta} u^{\alpha} & =-R_{\mu \alpha \nu}^{\beta} u_{\beta} u^{\alpha} \\ & =-R_{\beta \mu \alpha \nu} u^{\beta} u^{\alpha} . \tag{E.463} \end{align*}Rμναβuβuα=Rμανβuβuα(E.463)=Rβμανuβuα.
(d) This follows from the symmetry properties of the terms and the results of the previous problem.
(e) Use the result from (d) for the first three terms. For the final term, note R β μ α ν = R μ β ν α R β μ α ν = R μ β ν α R_(beta mu alpha nu)=R_(mu beta nu alpha)R_{\beta \mu \alpha \nu}=R_{\mu \beta \nu \alpha}Rβμαν=Rμβνα. The contraction with g μ ν g μ ν g^(mu nu)g^{\mu \nu}gμν then turns the Riemann tensor into the Ricci tensor.
(50.3) As in the text for the chapter, we substitute R μ ν u μ u ν = R μ ν u μ u ν = R_(mu nu)u^(mu)u^(nu)=R_{\mu \nu} u^{\mu} u^{\nu}=Rμνuμuν= 8 π ( T μ ν 1 2 g μ ν T ) u μ u ν 8 π T μ ν 1 2 g μ ν T u μ u ν 8pi(T_(mu nu)-(1)/(2)g_(mu nu)T)u^(mu)u^(nu)8 \pi\left(T_{\mu \nu}-\frac{1}{2} g_{\mu \nu} T\right) u^{\mu} u^{\nu}8π(Tμν12gμνT)uμuν. Considering the strong energy condition then guarantees that the right-hand side of the equation is negative, and so the congruence is forced to contract under the influence of gravity.
(D.1) We can embed a parabolic 2-surface in R 3 R 3 R^(3)\mathbb{R}^{3}R3 using method II. The parabolic surface in Euclidean 3-space is described by
(E.464) Z = a 2 ( X 2 + Y 2 ) (E.464) Z = a 2 X 2 + Y 2 {:(E.464)Z=(a)/(2)(X^(2)+Y^(2)):}\begin{equation*} Z=\frac{a}{2}\left(X^{2}+Y^{2}\right) \tag{E.464} \end{equation*}(E.464)Z=a2(X2+Y2)
so we have
d Z = a X d X + a Y d Y d Z = a X d X + a Y d Y dZ=aXdX+aYdY\mathrm{d} Z=a X \mathrm{~d} X+a Y \mathrm{~d} YdZ=aX dX+aY dY
(E.465)
We write
d s 2 = d X 2 + d Y 2 + a 2 ( X d X + Y d Y ) 2 d s 2 = d X 2 + d Y 2 + a 2 ( X d X + Y d Y ) 2 ds^(2)=dX^(2)+dY^(2)+a^(2)(XdX+YdY)^(2)\mathrm{d} s^{2}=\mathrm{d} X^{2}+\mathrm{d} Y^{2}+a^{2}(X \mathrm{~d} X+Y \mathrm{~d} Y)^{2}ds2=dX2+dY2+a2(X dX+Y dY)2
and using the substitutions we have before: ( X , Y ) = ( X , Y ) = (X,Y)=(X, Y)=(X,Y)= ( r cos θ , r sin θ ) ( r cos θ , r sin θ ) (r cos theta,r sin theta)(r \cos \theta, r \sin \theta)(rcosθ,rsinθ), we find
(E.467) d s 2 = ( 1 + a 2 r 2 ) d r 2 + r 2 d θ 2 (E.467) d s 2 = 1 + a 2 r 2 d r 2 + r 2 d θ 2 {:(E.467)ds^(2)=(1+a^(2)r^(2))dr^(2)+r^(2)dtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=\left(1+a^{2} r^{2}\right) \mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2} \tag{E.467} \end{equation*}(E.467)ds2=(1+a2r2)dr2+r2 dθ2
(D.2) The equation of the surface in R 4 R 4 R^(4)\mathbb{R}^{4}R4 is given by X 2 + Y 2 + X 2 + Y 2 + X^(2)+Y^(2)+X^{2}+Y^{2}+X2+Y2+ Z 2 + W 2 = 1 Z 2 + W 2 = 1 Z^(2)+W^(2)=1Z^{2}+W^{2}=1Z2+W2=1. If we use spherical coordinates for X X XXX, Y Y YYY and Z Z ZZZ we have X = r sin θ cos ϕ , Y = r sin θ sin ϕ X = r sin θ cos ϕ , Y = r sin θ sin ϕ X=r sin theta cos phi,Y=r sin theta sin phiX=r \sin \theta \cos \phi, Y=r \sin \theta \sin \phiX=rsinθcosϕ,Y=rsinθsinϕ, Z = r cos θ Z = r cos θ Z=r cos thetaZ=r \cos \thetaZ=rcosθ and we find that W 2 = 1 r 2 W 2 = 1 r 2 W^(2)=1-r^(2)W^{2}=1-r^{2}W2=1r2 and W d W = r d r W d W = r d r WdW=-rdrW \mathrm{~d} W=-r \mathrm{~d} rW dW=r dr. We can use this to eliminate W W WWW from the line element
d s 2 = d X 2 + d Y 2 + d Z 2 + d W 2 = d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) + ( r d r ) 2 1 r 2 = d r 2 1 r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = d r 2 1 r 2 + r 2 d Ω 2 2 d s 2 = d X 2 + d Y 2 + d Z 2 + d W 2 = d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 + ( r d r ) 2 1 r 2 = d r 2 1 r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 d s 2 = d r 2 1 r 2 + r 2 d Ω 2 2 {:[ds^(2)=dX^(2)+dY^(2)+dZ^(2)+dW^(2)],[=dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))+((-r(d)r)^(2))/(1-r^(2))],[=(dr^(2))/(1-r^(2))+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2))],[ds^(2)=(dr^(2))/(1-r^(2))+r^(2)dOmega_(2)^(2)]:}\begin{aligned} \mathrm{d} s^{2} & =\mathrm{d} X^{2}+\mathrm{d} Y^{2}+\mathrm{d} Z^{2}+\mathrm{d} W^{2} \\ & =\mathrm{d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right)+\frac{(-r \mathrm{~d} r)^{2}}{1-r^{2}} \\ & =\frac{\mathrm{d} r^{2}}{1-r^{2}}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \\ & \mathrm{d} s^{2}=\frac{\mathrm{d} r^{2}}{1-r^{2}}+r^{2} \mathrm{~d} \Omega_{2}^{2} \end{aligned}ds2=dX2+dY2+dZ2+dW2=dr2+r2( dθ2+sin2θ dϕ2)+(r dr)21r2=dr21r2+r2( dθ2+sin2θ dϕ2)ds2=dr21r2+r2 dΩ22
or
with d Ω 2 2 = d θ 2 + sin 2 θ d ϕ 2 d Ω 2 2 = d θ 2 + sin 2 θ d ϕ 2 dOmega_(2)^(2)=dtheta^(2)+sin^(2)thetadphi^(2)\mathrm{d} \Omega_{2}^{2}=\mathrm{d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}dΩ22=dθ2+sin2θ dϕ2. This looks very much like the D = 2 D = 2 D=2D=2D=2 version, with the replacement of d ϕ d ϕ dphi\mathrm{d} \phidϕ with d Ω 2 d Ω 2 dOmega_(2)\mathrm{d} \Omega_{2}dΩ2. If we now write r = sin ψ r = sin ψ r=sin psir=\sin \psir=sinψ we obtain
(E.470) d s 2 = d ψ 2 + sin 2 ψ ( d θ 2 + sin 2 θ d ϕ 2 ) (E.470) d s 2 = d ψ 2 + sin 2 ψ d θ 2 + sin 2 θ d ϕ 2 {:(E.470)ds^(2)=dpsi^(2)+sin^(2)psi((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} \psi^{2}+\sin ^{2} \psi\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{E.470} \end{equation*}(E.470)ds2=dψ2+sin2ψ( dθ2+sin2θ dϕ2)
and we guess that
(E.471) d Ω 3 2 = d ψ 2 + sin 2 ψ d Ω 2 2 (E.471) d Ω 3 2 = d ψ 2 + sin 2 ψ d Ω 2 2 {:(E.471)dOmega_(3)^(2)=dpsi^(2)+sin^(2)psidOmega_(2)^(2):}\begin{equation*} \mathrm{d} \Omega_{3}^{2}=\mathrm{d} \psi^{2}+\sin ^{2} \psi \mathrm{~d} \Omega_{2}^{2} \tag{E.471} \end{equation*}(E.471)dΩ32=dψ2+sin2ψ dΩ22
(D.3) The equation for the surface may be expressed through the parametric equations
X = ( c + a cos v ) cos u , Y = ( c + a cos v ) sin u , (E.472) Z = a sin v . X = ( c + a cos v ) cos u , Y = ( c + a cos v ) sin u , (E.472) Z = a sin v . {:[X=(c+a cos v)cos u","],[Y=(c+a cos v)sin u","],[(E.472)Z=a sin v.]:}\begin{align*} & X=(c+a \cos v) \cos u, \\ & Y=(c+a \cos v) \sin u, \\ & Z=a \sin v . \tag{E.472} \end{align*}X=(c+acosv)cosu,Y=(c+acosv)sinu,(E.472)Z=asinv.
We can use the method-I rules to determine the induced metric
X u = ( c + a cos v ) sin u , X v = a sin v cos u , Y u = ( c + a cos v ) cos u , Y Y = a sin v sin u , Z u = 0 , Z v = a cos v , X u = ( c + a cos v ) sin u ,      X v = a sin v cos u , Y u = ( c + a cos v ) cos u ,      Y Y = a sin v sin u , Z u = 0 ,      Z v = a cos v , {:[(del X)/(del u)=-(c+a cos v)sin u",",(del X)/(del v)=-a sin v cos u","],[(del Y)/(del u)=(c+a cos v)cos u",",(del Y)/(del Y)=-a sin v sin u","],[(del Z)/(del u)=0",",(del Z)/(del v)=a cos v","]:}\begin{array}{ll} \frac{\partial X}{\partial u}=-(c+a \cos v) \sin u, & \frac{\partial X}{\partial v}=-a \sin v \cos u, \\ \frac{\partial Y}{\partial u}=(c+a \cos v) \cos u, & \frac{\partial Y}{\partial Y}=-a \sin v \sin u, \\ \frac{\partial Z}{\partial u}=0, & \frac{\partial Z}{\partial v}=a \cos v, \end{array}Xu=(c+acosv)sinu,Xv=asinvcosu,Yu=(c+acosv)cosu,YY=asinvsinu,Zu=0,Zv=acosv,
from which we obtain the induced metric
(E.474) d s 2 = ( c + a cos v ) 2 d u 2 + a d v 2 (E.474) d s 2 = ( c + a cos v ) 2 d u 2 + a d v 2 {:(E.474)ds^(2)=(c+a cos v)^(2)du^(2)+adv^(2):}\begin{equation*} \mathrm{d} s^{2}=(c+a \cos v)^{2} \mathrm{~d} u^{2}+a \mathrm{~d} v^{2} \tag{E.474} \end{equation*}(E.474)ds2=(c+acosv)2 du2+a dv2
(D.5) Taking derivatives, we find non-zero components g x x = g x x = g_(xx)=g_{x x}=gxx= 1 and g t t = x 2 g t t = x 2 g_(tt)=-x^(2)g_{t t}=-x^{2}gtt=x2, giving the metric d s 2 = x 2 d t 2 + d x 2 d s 2 = x 2 d t 2 + d x 2 ds^(2)=-x^(2)dt^(2)+dx^(2)\mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2}ds2=x2 dt2+dx2.